68 MATHEMATICS
or, a+ 2d = 600
and a+ 6d = 700
Solving these equations, we get d=25 and a=550.
Therefore, production of TV sets in the first year is 550.
Gi) Now Gy = 4+ 9d =5504+9 x 25=775
So, production of TV sets in the 10th year is 775.
- 7
(iii) Also, S,= 5 [2x 550+ (7-1) x25] .
7 :
= 3 [1100 + 150] =4375 .
Thus, the total production of TV sets in first 7 years is4375. -
EXERCISE 5.3. - |
1. Find the sum of the following APs: o ; :
(i) 2,7, 12,...,to 10 terms. ' di) -37,-33,-29,..., to 12 terms.
(ui) 0.6, 1.7, 2.8,..., to 100 terms. (iv) 1 i I ,--., to 11 terms.
. 15 12 10
2. Find the sums given below: ~
f :
1: ,
a) 7+ 105 hiss) asa . (i) 344324+304+...4+10
(iii) -5 +(-8)+(-11) +... + (230)
3. InanAP: ,
@) givena=5,d=3,a,=50, findnandS .
(ii) given a =7,a,,=35, find dand §,,.
(ii) given a,, = 37, d=3, find a and S.,.
(iv) givena,=15,S,,= 125, findd anda,,.
(v) givend=5,S,=75, find a and a,.
(vi) givena=2, d=8, S =90, findn anda,
(vil) givena=8,a,=62,S =210, find n and d.
(viii) givena =4,d=2, S =-14, find n and a.
(x) givena=3,n=8,S = 192, findd.
(x) given/=28, S = 144, and there are total 9 terms. Find a.
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ARITHMETIC PROGRESSIONS 69

4, How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?

5, The first term of an APis 5, the last term is 45 and the sum is 400. Find the number of terms
and the common difference.

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference
is 9, how many terms are there and what is their sum?

7. Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively.

9, If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of
first n terms.

10. Show that a,,a,,...,a,,...formanAP where a, is defined as below :

@) a, =3+44n (i) a,=9-5n ; :
Also find the sum of the first 15 terms in each case. . ,

11. If the sum of the first n terms of an AP is 4n - n?, what is the first term (that is S,)? What
is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and
the nth terms. , ° -

12. Find the sum of the first 40 positive integers divisible by 6.

13. Find the sum of the first 15 multiples of 8.

14. Find the sum of the odd numbers between 0 and 50.

15. A contract on construction job specifies a penalty for delay of completion beyond a
certain date as follows: % 200 for the first day, ¥ 250 for the second day, % 300 for the third
day, etc., the penalty for each succeeding day being = 50 more than for the preceding day.
How much money the contractor has to pay as penalty, if he has delayed the work by 30
days? °

16. A sum of % 700 is to be used to give seven cash prizes to students of a school for their
overall academic performance. If each prize is = 20 less than its preceding prize, find the
value of each of the prizes.

17. Ina school, students thought of planting trees in and around the school to reduce air
pollution. It was decided that the number of trees, that each section of each class will
plant, will be the same as the class, in which they are studying, e.g., a section of Class I
will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are
three sections of each class. How many trees will be planted by the students?

18. A spiral is made up of successive semicircles, with centres alternately at A and B,
starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,... as shown in
Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive

. 22
semicircles? (Take n = 7?
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70 MATHEMATICS
I, ,
I, :
Fig. 5.4 ;
[Hint : Length of successive semicircles is Lt, L, Lye. with centres at A, B, A, B,...,
respectively. ] .

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row,
18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed
and how many logs are in the top row? a "

APIS TASA SAS LALA LLL J
Lip ipl iphiPh ipl flip iphip_iphiPhiplhip ipl phi php
Aofofofofofofofofofofofofofeofofofofofe)
-_~ " Fig. 5.5

20. Ina potato race, a bucket is placed at the starting point, which is 5 m from the first potato,
and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the
line (see Fig. 5.6).

5m a) m 3m e e . ° e °
Fig. 5.6

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops
it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and
she continues in the same way until all the potatoes are in the bucket. What is the total
distance the competitor has to run?

[Hint : To pick up the first potato and the second potato, the total distance (in metres)
run by acompetitor is 2x5+2x(5+3)]

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ARITHMETIC PROGRESSIONS 71
EXERCISE 5.4 (Optional)*
1. Which term of the AP: 121, 117, 113,..., is
its first negative term?
{Hint : Find n for a, < 0] 7S cn
2. The sum of the third and the seventh terms ' I
of an AP is 6 and their product is 8. Find ! \
the sum of first sixteen terms of the AP. ! \
! 1
3. A ladder has rungs 25 cm apart. ' 1
(see Fig. 5.7). The rungs decrease ! ' ot m :
uniformly in length from 45 cm at the 2
bottom to 25 cm at the top. If the top and
1] , S
the bottom rungs are 25 m apart, what is cope
the length of the wood required for the . . ; 4 cm
rungs? " 45 cm cee
{Hint : Number of rungs = ~ +1] / Fig. 5.7
4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value
of x such that the sum of the numbers of the houses preceding the house numbered x is
equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint:S,_,=S,,7 S.J . °
5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and
built of solid concrete." .
1 1
Each step has a rise of qm and a tread of 5m. (see Fig. 5.8). Calculate the total volume
of concrete required to build the terrace.
. 1 1
{Hint : Volume of concrete required to build the first step = 4 x 5 x 50m' ]
7 5
1 an oF
q m~__y, go
Fig. 5.8
* These exercises are not from the examination point of view.
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72 MATHEMATICS
5.5 Summary
In this chapter, you have studied the following points :

1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by
adding a fixed number d to the preceding term, except the first term. The fixed number d
is called the common difference.

The general form of anAPisa, a+d, a+2d, a+3d,...

2. A given list of numbers a,, a,, a,, .. . is an AP, if the differences a, - a,, a, - a,
A,- y+. give the same value, 1.e., if 4,,,-4 is the same for different values of k.

3. In an AP with first term a and common difference d, the nth term (or the general term) is
givenby a, =a+(n-l)d.

4, The sum of the first m terms of an AP is given by :

S= 5l24 + (nd) .
5. If J is the last term of the finite AP, say the nth term, then the sum of all terms of the AP
is given by: - ' ° -
S= (atl
= 5 (a +1)
- A NOTE TO THE READER
. a + c . » .
If a, b, c are in AP, then b = 3 and b is called the arithmetic
mean of a and c.
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a . il
talk : arf
1062CHO6
6.1 Introduction .
You are familiar with triangles and many of their properties from your earlier classes.
In Class IX, you have studied congruence of triangles in detail. Recall that two figures
are said to be congruent, if they have the same shape and the same size. In this
chapter, we shall study about those figures which have the same shape but not necessarily
the same size. Two figures having the same shape (and not necessarily the same size)
are called similar figures. In particular, we shall discuss the similarity of triangles and
apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
Can you guess how heights of mountains (say Mount Everest) or distances of
some long distant objects (say moon) have been found out? Do you think these have
_- - WAH WAHT =
AAH! Soon Iwill nS
It is so easy reach the moon,
to measure the} |
height of the Sf
mountain, Uh
ry
coN Aas
BNL? Et)
. S9 ooo - tn
¥ . a: ----
"Os - J [st vie
Va br pe
ey, V Chey s TW,
eS | | rT
eT i
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74 MATHEMATICS
been measured directly with the help of a measuring tape? In fact, all these heights
and distances have been found out using the idea of indirect measurements, which is
based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3
and also Chapters 8 and 9 of this book).

6.2 Similar Figures

In Class IX, you have seen that all circles with the same radii are congruent, all
squares with the same side lengths are congruent and all equilateral triangles with the
same side lengths are congruent. "

Now consider any two (or more)
circles [see Fig. 6.1 (i)]. Are they
congruent? Since all of them do not
have the same radius, they are not
congruent to each other. Note that . Ce .
some are congruent and some are not, - @
but all of them have the same shape. _ °
So they all are, what we call, similar. - :

Two similar figures have the same_ |. ,

shape but not necessarily the same

size. Therefore, all circles are similar. . . | |
What about two (or more) squares or

two (or more) equilateral triangles : (ii)
[see Fig. 6.1 (ii) and (iii)]? As observed +

in the case of circles, here also all

squares are similar and all equilateral

triangles are similar. /\

From the above, we can say
that all congruent figures are (iii)
similar but the similar figures need
not be congruent. Fig. 6.1

Can a circle and a square be P S
similar? Can a triangle and a square A D
be similar? These questions can be | \
answered by just looking at the
figures (see Fig. 6.1). Evidently B c Q
these figures are not similar. (Why?) R

Fig. 6.2
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TRIANGLES 75

What can you say about the two quadrilaterals ABCD and PQRS
(see Fig 6.2)?Are they similar? These figures appear to be similar but we cannot be
certain about it.Therefore, we must have some definition of similarity of figures and
based on this definition some rules to decide whether the two given figures are similar
or not. For this, let us look at the photographs given in Fig. 6.3:

« Lan at a . Lo .
nef a} | | aod sh, Ay
ie Crh) [alee |jpd owl
CATR NOC acre ytty | bBiaseernty
Fig. 6.3 :

You will at once say that they are the photographs of the same monument
(Taj Mahal) but are in different sizes. Would you say that the three photographs are
similar? Yes,they are. 7 ce a

What can you say about the two photographs of the same size of the same
person one at the age of 10 years and the other at the age of 40 years? Are these
photographs similar? These photographs are of the same size but certainly they are
not of the same shape. So, they are not similar.

What does the photographer do when she prints photographs of different sizes
from the same negative? You must have heard about the stamp size, passport size and
postcard size photographs. She generally takes a photograph on a small size film, say
of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we
consider any line segment in the smaller photograph (figure), its corresponding line

45 55

segment in the bigger photograph (figure) will be 35 [0 =) of that of the line segment.
This really means that every line segment of the smaller photograph is enlarged
(increased) in the ratio 35:45 (or 35:55), It can also be said that every line segment
of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further,
if you consider inclinations (or angles) between any pair of corresponding line segments
in the two photographs of different sizes, you shall see that these inclinations(or angles)
are always equal. This is the essence of the similarity of two figures and in particular
of two polygons. We say that:

Two polygons of the same number of sides are similar, if (i) their
corresponding angles are equal and (ii) their corresponding sides are in the
same ratio (or proportion).

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76 MATHEMATICS

Note that the same ratio of the corresponding sides is referred to as the scale
factor (or the Representative Fraction) for the polygons. You must have heard that
world maps (i.e., global maps) and blue prints for the construction of a building are
prepared using a suitable scale factor and observing certain conventions.

In order to understand similarity of figures more clearly, let us perform the following
activity:

Activity 1: Place a lighted bulb at a Oo

point O on the ceiling and directly below

it a table in your classroom. Let us cut a Ss R= :

polygon, say a quadrilateral ABCD, from "Zs - x

a plane cardboard and place this AN ,

cardboard parallel to the ground between i (i .

the lighted bulb and the table. Then a a if '\

shadow of ABCD is cast on the table. hiss Hf WW

Mark the outline of this shadow as Lak A ij mee \ B

A'B'C'D' (see Fig.6.4). ; As. -

Note that the quadrilateral A'B'C'D' is ° aK io oc iN
so : ,) ; i a / - BY »

an enlargement (or magnification) of the & j Yy LN. 2

quadrilateral ABCD. This is because of: | V yy LY Jaa

the property of light that light propogates v {n. TRA TN are <= |

in a straight line. You may also note that. oe Fig. 6.4

A' lies on ray OA, B' lies on ray OB, C' » .

lies on OC and D' lies on OD. Thus, quadrilaterals A'B'C'D' and ABCD are of the

same shape but of different sizes. .

So, quadrilateral A'B'C'D' is similiar to quadrilateral ABCD. We can also say
that quadrilateral ABCD is similar to the quadrilateral A'B'C'D'.

Here, you can also note that vertex A' corresponds to vertex A, vertex B'
corresponds to vertex B, vertex C' corresponds to vertex C and vertex D' corresponds
to vertex D. Symbolically, these correspondences are represented as A' > A, B' OB,
C' © C and D' © D. By actually measuring the angles and the sides of the two
quadrilaterals, you may verify that

@) ZA=ZA',ZB=2ZB',Z2C=2C,ZD=ZD' and

(ii) AB BC CD DA

ny - = --_ = --_ = --_
AB BC CD' DA'

This again emphasises that two polygons of the same number of sides are
similar, if (@) all the corresponding angles are equal and (ii) all the corresponding
sides are in the same ratio (or proportion).

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TRIANGLES 77
From the above, you can easily say that quadrilaterals ABCD and PQRS of
Fig. 6.5 are similar.
4.8 cm | R
5 70°
\/ss°
2.4 Cc
D = 70°) 5.0 cm
35° 4.2 cm "
2.5 cm
2.1 cm 105° 3
105° 100 100 .
1.5 cm P. 3.0 cm ~Q..
Fig.6.5 0 | -
Remark : You can verify that if one polygon is similar to another polygon and this
second polygon is similar to a third polygon, then the first polygon is similar to the third
polygon.
You may note that in the two quadrilaterals (a square and a rectangle) of
Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the
same ratio. fc ) .
D 3 cm Cc . 3.5 cm
[J [J S R
3 cm "ot 3 cm 3cm 3 cm
7 : TI P
A 3 cm B 3.5 cm Q
_ Fig. 6.6
So, the two quadrilaterals are not similar. Similarly, you may note that in the two
quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the
same ratio, but their corresponding angles are not equal. Again, the two polygons
(quadrilaterals) are not similar.
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78 MATHEMATICS
S
4.2 cm
D 2.1em_¢ 4.2 cm R
2.1..em 2.1.em
P 4.2 cm
4.2 cm
A Ziem ®
Q -
Fig. 6.7 ;
Thus, either of the above two conditions (i) and (ii) of similarity of two
polygons is not sufficient for them to be similar. :
EXERCISE 6.1. Do
1. Fill in the blanks using the correct word given in brackets : °
(i) All circles are _______. (congruent, similar) .

(i) All squares are _______. (similar, congruent)

(ai) All_____ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding
angles are <_~____ and (b) their corresponding sides are _______. (equal,
proportional) ) ,

2. Give two different examples of pair of ,
(i) similar figures. (ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:
a ; D 3cm C
S_15cm_R 3cm 3cm
1.5 of fs cm
P 1.5 cm Q A 3cm B
Fig. 6.8
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TRIANGLES 79
6.3 Similarity of Triangles
What can you say about the similarity of two triangles?

You may recall that triangle is also a polygon. So, we can state the same conditions
for the similarity of two triangles. That is:

Two triangles are similiar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

Note that if corresponding angles of two
triangles are equal, then they are known as
equiangular triangles. A famous Greek Ke
mathematician Thales gave an important truth relating iG a .
to two equiangular triangles which is as follows: . at oY,

The ratio of any two corresponding sides in Ces ff ;
two equiangular triangles is always the same. - J Nn wy

It is believed that he had used a result called : , "Dp
the Basic Proportionality Theorem (now known as | -
the Thales Theorem) for the same. ; , 4 mares c

640 - 546 B.C.

To understand the Basic Proportionality ( )
Theorem, let us perform the following activity: .
Activity 2 : Draw any angle XAY and on its one
arm AX, mark points (say five points) P, Q, D, R and A E Cc ¥
B such that AP = PQ = QD = DR = RB- P

Now, through B, draw any line intersecting arm Q D
AY at C (see Fig. 6.9). R 3

Also, through the point D, draw a line parallel x
to BC to intersect AC at E. Do you observe from

tructions that "> ==?M AE and neo
your constructions that > = ? Measure an
EC. What about =? Ob hat A= is al Ito =. Th h
. What about EC. serve that EC 18 also equa to 7; Thus, you can see that
; AD_AE.._., ue ,
in A ABC, DE|| BC and DB = EC' Is it a coincidence? No, it is due to the following
theorem (known as the Basic Proportionality Theorem):
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80 MATHEMATICS
Theorem 6.1 : [f a line is drawn parallel to one side of a triangle to intersect the
other two sides in distinct points, the other two sides are divided in the same
ratio. A
Proof : We are given a triangle ABC in which a line M
parallel to side BC intersects other two sides AB and N,
AC at D and E respectively (see Fig. 6.10). D E
Ww dt that AD = AE
e need to prove that D> EC' B
Cc
Let us join BE and CD and then draw DM L AC and
EN 1 AB. Fig. 6.10
1 ; 1 :
Now, area of A ADE (= 2 base x height) = 2 AD x EN. .
Recall from Class IX, that area of A ADE is denoted as ar(ADE). .
1 a
So, ar(ADE) = 5 AD x EN * ;
_. 1 .
Similarly, ar(BDE) = 2 DB x EN, _
ar(ADE) = 5 AE x DM and ar(DEC) = 5 EC x DM.
! AD x EN
Theref ar(ADE) 5" * AD a)
erefore, -r OE OT Er
ar(BDE) 1 pgxEN PB
ao 2
I AE x DM
i ar(ADE) 5 /*~* AE Q)
an 2
» ar(DEC) 1 ECxpM EC
2
Note that A BDE and DEC are on the same base DE and between the same parallels
BC and DE.
So, ar(BDE) = ar(DEC) (3)
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TRIANGLES 81
Therefore, from (1), (2) and (3), we have :

AD _ AE

DB EC

Is the converse of this theorem also true (For the meaning of converse, see
Appendix 1)? To examine this, let us perform the following activity:

Activity 3: Draw an angle XAY on your B, A

notebook and on ray AX, mark points B,, B,, B B, C,

BoB and such that AB, = B,B, = B,B, = B, 2 C,
374 7 "4" B C; :

Similarly, on ray AY, mark points X Cc,
C,, C,, C,, C, and C such that AC, = C,C, = : Cc
CC, = C,C, = C,C. Then join B,C, and BC © ; Se
(see Fig. 6.11). , . oe Y

. - Fig. 6.11
AB, AC, . 1 :
Note that B,B = Gc: (Each equal to 4
You can also see that lines B,C, and BC are parallel to each other, i.c.,
BC, || BC. (1)
Similarly, by joining B,C,, B,C, and B,C,, you can see that:

f .

BB cc \ 3) mM BGI

ABs AC, - 3) d BY, || BC 3

BB Gc. 2) @™ BGI @)
~ AB, _ ACS (=) and 3,¢, BC )

B,B. c,c (1) O° Pas

From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a
triangle in the same ratio, then the line is parallel to the third side.

You can repeat this activity by drawing any angle XAY of different measure and
taking any number of equal parts on arms AX and AY . Each time, you will arrive at
the same result. Thus, we obtain the following theorem, which is the converse of
Theorem 6.1:

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82 MATHEMATICS
Theorem 6.2 : /f a line divides any two sides of a A
triangle in the same ratio, then the line is parallel
to the third side.
This theorem can be proved by taking a line DE such D E',
= E
h AD _ AE d ing that DEi Hel .
that DB. EC 22 assuming that DE is not paralle
to BC (see Fig. 6.12). B
Cc
If DE is not parallel to BC, draw a line DE' . ;
parallel to BC. Fig. 6.12
S AD _ AE Why ?
0, DB _ E'C ( y 2) ;
Theref AE AE why? | ot
erefore, FC" EC ( y ?) / a .
Adding 1 to both sides of above, you can see that E and E' must coincide.
(Why ?) oo -
Let us take some examples to illustrate the use of the above theorems.
Example 1 : Ifa line intersects sides AB and AC of a A ABC at D and E respectively
d is parallel to BC, prove that "2 = "© (see Fig. 6.13
and is parallel to BC, Prove at 7B = AC (see Fig. 6.13).
MN .
Solution : DE || BC (Given)
S ADL AE Thi 6.1
0, DB = kc (Theorem 6.1)
. A
DB _ EC
" - AD AE
DB 1 EC 1
Or, . "AD + = AE + D E
. AB _ AC
° AD AE
B
5 AD _ AE c
° AB AC Fig. 6.13
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TRIANGLES 83
Example 2 : ABCD is a trapezium with AB || DC. E A B
and F are points on non-parallel sides AD and BC
respectively such that EF is parallel to AB E F
Fig. 6.14). Show th AE. BF
(see Fig. 6.14). Show that ED FC' 5
Cc
Solution : Let us join AC to intersect EF at G .
(see Fig. 6.15). Fig. 6.14
AB || DC and EF || AB (Given) '
So, EF||DC (Lines parallel to the same line are ;
parallel to each other)
A B
EG|| DC (As EF || DC) ok F
go, "B= AS 6.1 1
0, ED = GC (Theorem 6.1) CO)
OD Cc
Similarly, from A CAB, . -
cG CF Fig. 6.15
AG BF...
| C) AG _ BF 2
rE \/ GC FC P 2)
Therefore, from (1) and (2), ©
AE _ BF
-_ ED FC
PS) pT s v
Example 3 : In Fig. 6.16, Re) = TR and Z PST =
Z PRQ. Prove that PQR is an isosceles triangle. Q R
: PS _ PT Fi
eo =~ tf, g. 6.16
Solution : It is given that SQ TR
So, ST || QR (Theorem 6.2)
Therefore, Z PST = ZPQR (Corresponding angles) (1)
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84 MATHEMATICS
Also, it is given that
ZPST= Z PRQ Q2)
So, Z PRQ = Z PQR [From (1) and (2)]
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle.
EXERCISE 6.2
1. InFig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). .
A , x
A
15m Aj em a \us cm
D E 7.2m E:
3cm ,
B 5.4m
B C . .
(i) | Gi) c
Fig. 6.17 | B
2. E and F are points on the sides PQ and PR M
respectively of a A PQR. For each of the following A Cc
cases, state whether EF || QR : x
(i) PE=3.9cm, EQ=3 cm, PF=3.6cmand FR=2.4cm D
(i) PE=4cm, QE=4.5 cm, PF=8 cmand RF=9cm Fig. 6.18
(iit) PQ=1.28 cm, PR = 2.56 cm, PE =0.18 cm and PF=0.36cm
3. In Fig. 6.18, if LM || CB and LN || CD, prove that
A
AM _ AN'
AB AD D
4, In Fig. 6.19, DE || AC and DF || AE. Prove that
BF BE
aa. B Cc
FE EC FUE
Fig. 6.19
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TRIANGLES 85

5. In Fig. 6.20, DE || OQ and DF || OR. Show that P
EF] QR.

6. In Fig. 6.21, A, B and C are points on OP, OQ and E F
OR respectively such that AB || PQ and AC || PR.

Show that BC |] QR.
7. Using Theorem 6.1, prove that a line drawn through
. : . . Q R
the mid-point of one side of a triangle parallel to .
another side bisects the third side. (Recall that you Fig. 6.20
have proved it in Class IX).

8. Using Theorem 6.2, prove that the line joining the P '
mid-points of any two sides of a triangle is parallel A
to the third side. (Recall that you have done it in
Class IX). .

9. ABCD is a trapezium in which AB || DC and its LO
diagonals intersect each other at the point O. Show 6B

nat 22 = £2. fe
that BO. DO . . Fig. 6.21
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
AO 5, hat ABCD i seri
BO. DO ow that is a trapezium. |
6.4 Criteria for Similarity of Triangles
In the previous section, we Stated that two triangles are similar, if (1) their corresponding
angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
That is, in A ABC and A DEF, if
G@) ZA=2D,ZB=ZE,2C=Z Fand
(ii) AB _BC _CA> hen th iangl imil Fig. 6.22
ll) .-- = 2 = 1" then the two triangles are similar (See Fig. 6. .
DE EF FD 6 (see Fig. 6.22)
A
- L™. |
B Cc E F
Fig. 6.22
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86 MATHEMATICS
Here, you can see that A corresponds to D, B corresponds to E and C
corresponds to F. Symbolically, we write the similarity of these two triangles as
'A ABC ~ A DEF' and read it as 'triangle ABC is similar to triangle DEF'. The
symbol '~' stands for 'is similar to'. Recall that you have used the symbol '=' for
"is congruent to' in Class IX.
It must be noted that as done in the case of congruency of two triangles, the
similarity of two triangles should also be expressed symbolically, using correct
correspondence of their vertices. For example, for the triangles ABC and DEF of
Fig. 6.22, we cannot write A ABC ~ A EDF or A ABC ~ A FED. However, we
can write A BAC ~ A EDF.

Now a natural question arises : For checking the similarity of two triangles, say
ABC and DEF, should we always look for all the equality relations of their corresponding
angles (ZA=2D, ZB=ZE, ZC= Z FP and all the equality relations of the ratios

| - (AB = BE- CA), stusexamine. Y ith
of their corresponding sides DE EF ED). Let us examine. 'ou may recall that
in Class IX, you have obtained some criteria for congruency of two triangles involving
only three pairs of corresponding parts (or elements) of the two triangles. Here also,
let us make an attempt to arrive at certain criteria for similarity of two triangles involving
relationship between less number of pairs of corresponding parts of the two triangles,
instead of all the six pairs of corresponding parts. For this, let us perform the following
activity: i ) .

ML .
Activity 4: Draw two line segments BC and EF of two different lengths, say 3 cm
and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC
and QCB of some measures, say, 60° and 40°. Also, at the points E and F, construct
angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).
- * R
5 S D
Q P
NA
poe 407\¢ EA 40(\F
3cm 5 cm
Fig. 6.23
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TRIANGLES 87
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each
other at D. In the two triangles ABC and DEF, you can see that
ZB=ZE,2C=ZFand Z A= Z D. That is, corresponding angles of these two
triangles are equal. What can you say about their corresponding sides ? Note that
BC - 2-06 What ab , 0B dro ing AB, DE, CA and FD
[Fs . a ve DE 2nd Ep ? On measuring AB, DE, an , you
will find that DE and Ep we also equal to 0.6 (or nearly equal to 0.6, if there is some
AB BC CA ..
error in the measurement). Thus, DE EF ED You can repeat this activity by
constructing several pairs of triangles having their corresponding angles equal. Every
time, you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles. .
Theorem 6.3 : /f in two triangles, corresponding angles are equal, then their
corresponding sides are in the same ratio (or proportion) and hence the two
triangles are similar. : D
This criterion is referred to as the AAA" _ oe o
(Angle-Angle-Angle) criterion of _ A °
similarity of two triangles. . :
This theorem can be proved by taking two P Q
triangles ABC and DEF such that F
ZA=ZD,ZB=ZEandZC=ZF. B cr
(see Fig. 6.24) f ) . Fig. 6.24
Cut DP = AB and DQ = AC and join PQ.
So, 'A ABC = A DPQ (Why 2)
This gives ZB=ZP =ZE and PQ|| EF (How?)
therel ~ DP DQ Why?
erefore, 7 PE = OF (Why?)
| AB _ AC Why?
ie., | | DE DE (Why?)
simitary, OE = B andeo AB = BC-AC
mary DE" BF ""'S° DE EF ODF'
Remark : If two angles of a triangle are respectively equal to two angles of another
triangle, then by the angle sum property of a triangle their third angles will also be
equal. Therefore, AAA similarity criterion can also be stated as follows:
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88 MATHEMATICS

If two angles of one triangle are respectively equal to two angles of another
triangle, then the two triangles are similar.

This may be referred to as the AA similarity criterion for two triangles.

You have seen above that if the three angles of one triangle are respectively
equal to the three angles of another triangle, then their corresponding sides are
proportional (i.e., in the same ratio). What about the converse of this statement? Is the
converse true? In other words, if the sides of a triangle are respectively proportional to
the sides of another triangle, is it true that their corresponding angles are equal? Let us
examine it through an activity : '
Activity 5 : Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm,
CA =8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25).

D ; :
A a |
8 em 4.5 cm 12 om -
3 em .
B 6cm c E 9 cm F
Fig. 6.25 .
\ AB BC_ CA 2
So,youhave: (== =-~=-- (each equal to =
o, you have ( y DE EF FD (each equal to 3 )

Now measure Z A, Z B, ZC, Z D, Z E and Z F. You will observe that
ZA=2D,ZB=ZEand Z C= ZF ie,, the corresponding angles of the two
triangles are equal.

You can repeat this activity by drawing several such triangles (having their sides
in the same ratio). Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles:

Theorem 6.4 : /f in two triangles, sides of one triangle are proportional to
(i.e., in the same ratio of ) the sides of the other triangle, then their corresponding
angles are equal and hence the two triangles are similiar.

This criterion is referred to as the SSS (Side-Side-Side) similarity criterion for
two triangles.

This theorem can be proved by taking two triangles ABC and DEF such that
AB _ BC _©A ey (see Fig. 6.26):

DE EF ED (< 1) (see Fig. 6.26):
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TRIANGLES 89
D
A
| P Q
B Cc E F
Fig. 6.26
Cut DP = AB and DQ = AC and join PQ.
It can b h Pe OS and PQ IEF (How?)
t can be seen that PE = OF an Q || (How?) ;
So, ZP=ZE and. ZQ=ZF °
DP DQ PQ °
Theref - =a SE
"Ns DE DF EF
5 DP _DQ_BC
° 'DE DF EF WY?)
So, Cc) BC = PQ (Why?)
Thus, A ABC = A DPQ (Why ?)
So, ZA=ZD, ZB=ZE and ZC=ZF _ (How?)
Remark : You may recall that either of the two conditions namely, (i) corresponding
angles are equal and (11) corresponding sides are in the same ratio is not sufficient for
two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can
now say that in case of similarity of the two triangles, it is not necessary to check both
the conditions as one condition implies the other.

Let us now recall the various criteria for congruency of two triangles learnt in
Class IX. You may observe that SSS similarity criterion can be compared with the SSS
congruency criterion. This suggests us to look for a similarity criterion comparable to
SAS congruency criterion of triangles. For this, let us perform an activity.

Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm, Z A = 50°,
AC =4cm, DE=3 cm, Z D = 50° and DF = 6 cm (see Fig.6.27).
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90 MATHEMATICS
D
A DS
\>
4cm 6 cm
2cem 3cm
B Cc E F
Fig. 6.27 .
H bserve that ~~ = "© (cach equal to ~) and Z A (included
ere, you may observe that DE ~ DE (each equal to 3 ) an (ine ude
between the sides AB and AC) = Z D (included between the sides DE and DF). That
is, one angle of a triangle is equal to one angle of another triangle and sides including
these angles are in the same ratio (i.e., proportion). Now let us measure 2 B, 2 C,
ZEand ZF ; oe .

You will find that 7B = 2 Eand 2 C= ZF Thatis, ZA=ZD,ZB=ZEand
ZC=ZFE So, by AAA similarity criterion, A ABC ~ A DEF. You may repeat this
activity by drawing several pairs of such triangles with one angle of a triangle equal to
one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar. It is due to the following criterion
of similarity of triangles: "

y & / -_ ;
Theorem 6.5 : /f one angle of a triangle is equal to one angle of the other
triangle and the sides including these angles are proportional, then the two
triangles are similar.

This criterion is referred to as
the SAS (Side-Angle-Side) D
similarity criterion for two
triangles. ' ; A

As before, this theorem can
be proved by taking two triangles Q
ABC and DEF such that P
AB _ AC B CE F
- == 1 ZA=ZD
DE DE (< 1) and
(see Fig. 6.28). Cut DP = AB, DQ Fig. 6.28
= AC and join PQ.

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TRIANGLES 91
Now, PQ || EF and A ABC = A DPQ (How ?)
So, ZA=ZD, ZB=ZPandZC=ZQ
Therefore, A ABC ~ A DEF (Why?)
We now take some examples to illustrate the use of these criteria.
Example 4 : In Fig. 6.29, if PQ || RS, prove that A POQ ~ A SOR.
R
P '
oO , 5
Q So
Fig. 6.29 - oo
Solution : PQRS = - / : (Given)
So, ZP=ZS - ° (Alternate angles)
and ZQ=ZR
Also, Z POQ = ZSOR |. (Vertically opposite angles)
Therefore, vA POQ ~ ASOR (AAA similarity criterion)
Pa) .
Example 5 : Observe Fig. 6.30 and then find Z P.
R
A.
. 613 7.6
8
yee
Ast -
B ; Cc P a Q
Fig. 6.30
Solution ; In A ABC and A PQR,
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92 MATHEMATICS
AB_38_1 BC_6_1 = CA_3V3_1
RQ 76 2 QP 12 2° be 63 2
_ --AB_BC_CA
That is, RQ QP PR
So, A ABC ~ A RQP (SSS similarity)
Therefore, ZC=ZP (Corresponding angles of similar triangles)
But ZC= 180°-ZA-ZB (Angle sum property)
= 180° - 80° - 60° = 40° a
So, ZP= 40° . : CC
Example 6 : In Fig. 6.31, . ",
A.
OA .OB= OC. OD.
Show that ZA=ZCandZB=ZD.. | is
Solution: OA.OB=OC.OD~° (Given) .P |
OA OD : Fig. 6.31 3
-_ = -__ ] . .
So. OC ~ OB moe
Also, we have io ZAOD= Z COB (Vertically opposite angles) (2)
Therefore, from (1) and (2), AAOD~ ACOB (SAS similarity criterion)
So, ZA=ZCandZD=2B
. (Corresponding angles of similar triangles)
Example 7: A girl of height 90 cmis A
walking away from the base of a rs
lamp-post at a speed of 1.2 m/s. If the lamp ay
is 3.6 m above the ground, find the length "
of her shadow after 4 seconds.
Solution : Let AB denote the lamp-post Cc
and CD the girl after walking for4seconds J} , on
away from the lamp-post (see Fig. 6.32). HE E
From the figure, you can see that DE is the B ; D
shadow of the girl. Let DE be x metres. Fig. 6.32
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TRIANGLES 93
Now, BD = 1.22mx4=4.8 m.
Note that in A ABE and A CDE,

ZB=ZD_~- (Each is of 90° because lamp-post
as well as the girl are standing
vertical to the ground)

and ZE=ZE (Same angle)
So, A ABE ~ A CDE (AA similarity criterion)
Theref BE _ AB
erefore, DE = CD |
; 48+x 3.6 90 90 0 9
1.¢., x. = 09 ; ( cm = 700 m= 0. m)
ie, AS+x=4x - a |
i.e., 3x= 4.8 a - _
i.e., x= 1.6 - -
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example 8 : In Fig. 6.33, CM and RN are : Q N P
respectively the medians of A ABC and -
A PQR. If A ABC ~ A PQR, prove that :" A
(i) AAMC ~ APNR ) :
.. €M _ AB
@) RN PQ Cc
B R
(iit) A CMB ~ ARNQ
Solution: G) |. AABC~ A PQR ig. 6.33 (Given)
"AB BC_CA
So, PQ = QR. RP (1)
and - ZA=ZP, ZB=ZQandZC=ZR (2)
But AB = 2 AM and PQ = 2 PN
(As CM and RN are medians)
2AM CA
So, fi 1), = =
0, from (1) PN RP
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94 MATHEMATICS
| AM _ CA ;
i.e., PN = RP (3)
Also, ZMAC= Z NPR [From (2)] (4)
So, from (3) and (4),
A AMC ~ A PNR (SAS similarity) (5)
li) Fi 5 Me 6
(ii) From (5), aN = RP 6)
B CA _ AB Fi 1
ut RP = PQ [From (1)] (7)
CM _ AB
Therefore, RN = PQ [From (6) and (7) (8)
Gii) Again, PQ = OR . . ; [From (1)]
Theref CM _ BOO [From (8)] (9
erefore, RN = OR rom (8)] (9)
CM AB 2BM
Also, = =
- RN °PQ 2QN
. C) CM .BM 10
Le., - RN = ON (10)
| CM _ BC _BM . ;
Le., 4 RN = OR ON [From (9) and (10)]
Therefore, "OO A CMB ~ A RNQ (SSS similarity)
[Note ; You can also prove part (iii) by following the same method as used for proving
part (i).]
EXERCISE 6.3
1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by
you for answering the question and also write the pairs of similar triangles in the symbolic
form :
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TRIANGLES 95
P
P
A a A g 5
D>
sf ad =
BO 40/\\c¢ Qh 40-NR Bz Q R
(i) (ii)
P :
D ;
L M
4 g DP 5 °
"TL \ 2
[\
M7 P E 5 F N -L Q - 10 R
iv .
(ii > ey) .
: dD P
a
7
sy, : iw 80 39
B CE - t F.
3 (v) > ) . (vi)
: Fig. 6.34
2. In Fig. 6.35, AODC ~ AOBA, Z BOC = 125°
and ZCDO=70°. Find ZDOC, ZDCO and -
ZOAB. + '
OX )125°
3. Diagonals AC and BD of a trapezium ABCD / x
" A B
with AB || DC intersect each other at the
point O. Using a similarity criterion for two Fig. 6.35
anoles. chow that 2A = 2B.
triangles, show that OC OD
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96 MATHEMATICS
T
4. InFi 636, = 2 dZ1=22.Sh K
» InFig. 6.36, O¢ PR 20 = Z 2. Show
that A PQS ~ ATQR.
5. S and T are points on sides PR and QR of
A PQR such that Z P= 2 RTS. Show that Q S R
ARPQ~ ARTS. Fig. 6.36
6. In Fig. 6.37, if A ABE = A ACD, show that A
A ADE ~ A ABC. °
7. InFig. 6.38, altitudes AD and CE of AABC E .
intersect each other at the point P. Show D 4 .
that: . .
(i) AAEP ~ ACDP .
(ii) AABD ~ ACBE °
es: . " B Cc
(iii) AAEP~ AADB mo
(iv) APDC~ ABEC , | _ Fig. 6.37
8. Eis a point on the side AD produced of a Cc
parallelogram ABCD and BE intersects CD »D
at F. Show that A ABE ~ A CFB. _ .
9. In Fig. 6.39, ABC and AMP are two right :
triangles, right angled at B and M
respectively. Prove that:
A o B
(i) AABC ~ AAMP E
CA BC' Fig. 6.38
© PA MP
: C
10. CD and GH are respectively the bisectors
of Z ACB and Z EGF such that D and H lie M
on sides AB and FE of AABC and A EFG
respectively. If AABC ~ A FEG, show that:
. CD AC
© GH FG
A P
(ii) ADCB ~ AHGE B
(iii) ADCA~ AHGF Fig. 6.39
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TRIANGLES 97
11. In Fig. 6.40, E is a point on side CB A
produced of an isosceles triangle ABC
with AB =AC. If AD 1 BC and EF 1 AC,
prove that AABD ~ A ECF, F
12. Sides AB and BC and median AD of a
triangle ABC are respectively propor- E-=- -§ D Cc
tional to sides PQ and QR and median
PM of A PQR (see Fig. 6.41). Show that Fig. 6.40
A ABC ~ APQR.
13. Disa point on the side BC of a triangle A P
ABC such that Z ADC = Z BAC. Show :
that CA? =CB.CD.
14. Sides AB and AC and median AD of a
triangle ABC are respectively B D C QM M R
proportional to sides PQ and PR ands. oe ,
median PM of another triangle PQR. - Fig. 6.41
Show that AABC ~ A PQR. 5 . - _
15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time
a tower casts a shadow 28 m long. Find the height of the tower.
16. If AD and PM are medians of triangles ABC and PQR, respectively where
AB_AD
AABC~A wey PQ PM
6.5 Summary "- :
In this chapter you have studied the following points :
1. Two figures having the same shape but not necessarily the same size are called similar
figures. .
2. All the congruent figures are similar but the converse is not true.
3. Two polygons of the same number of sides are similar, if (1) their corresponding angles
are equal and (11) their corresponding sides are in the same ratio (i.e., proportion).
4. Ifa line is drawn parallel to one side of a triangle to intersect the other two sides in
distinct points, then the other two sides are divided in the same ratio.
5. Ifa line divides any two sides of a triangle in the same ratio, then the line is parallel to the
third side.
6. Ifintwo triangles, corresponding angles are equal, then their corresponding sides are in
the same ratio and hence the two triangles are similar (AAA similarity criterion).
7. Ifin two triangles, two angles of one triangle are respectively equal to the two angles of
the other triangle, then the two triangles are similar (AA similarity criterion).
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98 MATHEMATICS

8. If in two triangles, corresponding sides are in the same ratio, then their corresponding
angles are equal and hence the triangles are similar (SSS similarity criterion).

9. If one angle of a triangle is equal to one angle of another triangle and the sides including
these angles are in the same ratio (proportional), then the triangles are similar
(SAS similarity criterion).

A NOTE TO THE READER

If in two right triangles, hypotenuse and one side of one triangle are
proportional to the hypotenuse and one side of the other triangle,
then the two triangles are similar. This may be referred to as the
RHS Similarity Criterion.

If you use this criterion in Example 2, Chapter 8, the proof will become
simpler. '

OO.
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br, | |
am! [pr
2 fH3" er
1. er
a)! ate ai
loe2cHo?
COORDINATE GEOMETRY 7
7.1 Introduction .
In Class IX, you have studied that to locate the position of a point on a plane, we
require a pair of coordinate axes. The distance of a point from the y-axis is called its
x-coordinate, or abscissa. The distance of a point from the x-axis is called its
y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form
(x, 0), and of a point on the y-axis are of the form (0, y). _

Here is a play for you. Draw a set of a pair of perpendicular axes on a graph
paper. Now plot the following points and join them as directed: Join the point A(4, 8) to
BG, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to 1(8, 6) to
J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to
form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points
(0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6). What picture have you got?

Also, you have seen that a linear equation in two variables of the form
ax + by + c = 0, (a, 6 are not simultaneously zero), when represented graphically,
gives a straight line. Further, in Chapter 2, you have seen the graph of
y=ax?+bx+c(a#0), isa parabola. In fact, coordinate geometry has been developed
as an algebraic tool for studying geometry of figures. It helps us to study geometry
using algebra, and understand algebra with the help of geometry. Because of this,
coordinate geometry is widely applied in various fields such as physics, engineering,
navigation, seismology and art!

In this chapter, you will learn how to find the distance between the two points
whose coordinates are given. You will also study how to find the coordinates of the
point which divides a line segment joining two given points in a given ratio.

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100 MATHEMATICS
7.2 Distance Formula North

: . Y
Let us consider the following situation:

A town B is located 36 km east and 15
km north of the town A. How would you find
the distance from town A to town B without B
actually measuring it. Let us see. This situation
can be represented graphically as shown in 15 km
Fig. 7.1. You may use the Pythagoras Theorem A East x

. oy 36 km ;
to calculate this distance.

Now, suppose two points lie on the x-axis. Fig. 7.1 ;
Can we find the distance between them? For Y
instance, consider two points A(4, 0) and B(6, 0) 0.8 -
in Fig. 7.2. The points A and B lie on the x-axis. - 8 D(0, 8) .

From the figure you can see that OA = 4 7 ~ .
units and OB = 6 units, - 6 .

Therefore, the distance of B from A, i.e., °
AB =OB-OA=6-4=2 units. - 4

So, if two points lie on the x-axis, we can ;
easily find the distance between them. . 1

Now, suppose we take two points lying on * B x
the y-axis. Can you find the distance between O| 1 2 3/4 5 oN '0

. - : ,
them. If the points C(0, 3) and D(O, 8) lie on the (4, 0) (6, 0)
y-axis, similarly we find that CD = 8 --3 =5 units Fig. 7.2
(see Fig. 7.2).

Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC =3 units, the distance of A from C, i.c., AC= [3+ 4? =5 units, Similarly, you can
find the distance of B from D = BD = 10 units.

Now, if we consider two points not lying on coordinate axis, can we find the
distance between them? Yes! We shall use Pythagoras theorem to do so. Let us see
anexample. ~

In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant. How do we use
Pythagoras theorem to find the distance between them? Let us draw PR and QS
perpendicular to the x-axis from P and Q respectively. Also, draw a perpendicular
from P on QS to meet QS at T. Then the coordinates of R and S are (4, 0) and (6, 0),
respectively. So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.

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CoorDINATE GEOMETRY 101
Therefore, QT = 2 units and PT = RS = 2 units. Y
Now, using the Pythagoras theorem, we 8 Q(6, 8)
have 7
PQ = PT? + QT' 6 / T
= 242728 | 46)
So, PQ = 2,/2 units 3
How will we find the distance between two 2 ,
points in two different quadrants? 1 R R
Consider the points P(6, 4) and Q(-5, -3) ~ Of 1 2 374 5 ORG 0)
(see Fig. 7.4). Draw QS perpendicular to the (4,0) ,
x-axis. Also draw a perpendicular PT from the * Fig. 7.3.
point P on QS (extended) to meet y-axis at the a,
point R. ; a _
8
7
" "Ir 0,4 .
\ ') 3
- 2
5 1
"A [0123456 >
5,0) ~
oo 72
65-3)
, " Fig. 7.4
Then PT = 11 units and QT = 7 units. (Why?)
Using the Pythagoras Theorem to the right triangle PTQ, we get
PQ= J1? +7? = J170 units.
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102 MATHEMATICS
Let us now find the distance between any two Y
points P(x,, y,) and Q(x,, y,). Draw PR and QS Qe V2)
perpendicular to the x-axis. A perpendicular from the r
point P on QS is drawn to meet it at the point A
T (see Fig. 7.5). (py)
Then, OR=x,,OS=x,. So, RS =x, -x, = PT.
Also, SQ=y,, ST=PR=y,. So, QT=y,-y,.
Now, applying the Pythagoras theorem in A PTQ, we get *6 R S x
PQ? = PT? + QT? Fis. 7.5
ig. 7.
= (x,- x, + 0,-y,)
Therefore, PQ=, I(x» -X y +(y.- y,) . .
Note that since distance is always non-negative, we take only the positive square
root. So, the distance between the points P(,, y,) and Q(x,, y,) is."
2-7, 2 .
PQ = y(x2- 4) +(%- IH) >
which is called the distance formula. ° .
Remarks : ,
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by
co" .
\ OP = yx? +y"
2. We can also write, PQ = ./ (x, - x, y +(y, - yp y . (Why?)
Example 1 : Do the points (3, 2), (-2, -3) and (2, 3) form a triangle? If so, name the
type of triangle formed. |
Solution : Let us apply the distance formula to find the distances PQ, QR and PR,
where P(3, 2), Q(-2, -3) and RQ, 3) are the given points. We have
PQ = (342) + (2+3) =/5? +5? = V50 = 7.07 (approx.)
QR = (-2-2)? +3 -3)? =fC4)? + C6)? = V52 = 7.21 (approx.)
PR = (3 - 2)? +(2-3) =VI + (1? = V2 = 1.41 (approx.)
Since the sum of any two of these distances is greater than the third distance, therefore,
the points P, Q and R form a triangle.
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CoorDINATE GEOMETRY 103
Also, PQ? + PR? = QR', by the converse of Pythagoras theorem, we have Z P = 90°.
Therefore, PQR is a right triangle.
Example 2 : Show that the points (1, 7), (4, 2), (-1, -1) and (- 4, 4) are the vertices
of a square.
Solution : Let AC, 7), B(4, 2), C(-1, -1) and D(- 4, 4) be the given points. One way
of showing that ABCD is a square is to use the property that all its sides should be
equal and both its digonals should also be equal. Now,

AB =,/(1- 4)?+ (7-2) = 9 + 25 = V34 .

BC =,(44+1)?+ (241)? = /2549 = V34

CD =,/(-14+ 4)? + C1- 4)? =/9 + 25 = V34 .

DA=/(1+4)? + (7-4)? = f2549 = V34 CO

AC = (+1? + (7+? = (4+ 64=V68 .

BD = (4+ 4) +(2-4) = 644+4=J68
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral
ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a
square. . .
Alternative Solution : We find 10
the four sides and one diagonal, say, 9 P| ttt tt tt
AC as above. Here AD? + DC? = . PT} | fy | tty
34 + 34 = 68 = AC'. Therefore, by 8
the converse of Pythagoras. 7 Pt tt tt Tt
theorem, Z D=90°. A quadrilateral 6 BRR
with all four sides equal and one
mcs asa Sonsed ®™ spt
is a square. . . 4 3 S|

st le |
Example 3 : Fig. 7.6 shows the | | ff fy} toi
arrangement of desks in a 2 |
classroom. Ashima, Bharti and 1 ttt ttt
Camella are seated at A(3, 1), | | EL ELL] Ly
B(6, 4) and C(8, 6) respectively. 123 45 6 7 8 9 10
Do you think they are seated in a Columns
line? Give reasons for your Fig. 7.6
answer.
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104 MATHEMATICS
Solution : Using the distance formula, we have

AB = (6-3)? +(4-D? =/9+9 = Vi8 =3V2

BC = (8 - 6)? +(6 - 4)? =./4 +4 = v8 = 22

AC = (8 - 3)? + (6-1) = 25 + 25 = V50 = 5V2
Since, AB + BC = 32 + 2/2 = 5/2 = AC, we can say that the points A, B and C
are collinear. Therefore, they are seated in a line. |
Example 4 : Find a relation between x and y such that the point (x , y) is equidistant
from the points (7, 1) and (3, 5).
Solution ; Let P(x, y) be equidistant from the points A(7, 1) and BG, 5).
We are given that AP = BP. So, AP? = BP? | Ce
1.¢., (x-7+(y-1P = («-3P +(y-5P
i.e., x -14x+49+y--2y+1=x-6x4+9+y? - 10y + 25
Le., X-y=2 ° . | :
which is the required relation. - . . Y
Remark : Note that the graph of the equation (8
x-y=2 isa line. From your earlier studies, 7
you know that a point which is equidistant ~ 6 BG, 5)
from A and B lies on the perpendicular 5
bisector of AB. Therefore, the graph of 4
x-y=2 is the perpendicular bisector of AB 3
(see Fig. 7.7). 2
Example 5: Finda point on the y-axis which 1 A(7, 1)
. . . . xX
is equidistant from the points A(6, 5) and O 134567
B(- 4, 3). .

\
Solution : We know that a point on the 9
y-axis is of the form (0, y). So, let the point +7 .
P(O, y) be equidistant from A and B. Then Fig. 7.7
(6-0 +G6-yY=C4-0P + B-yy
1.¢., 364+25+y -10y= 164+9+y-6y
1.¢., 4y = 36
Le., y=9
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CoorDINATE GEOMETRY 105
So, the required point is (0, 9).
Let us check our solution: AP = ,/(6 - 0)? + (5-9)? =./36+16 = V52
BP = (- 4-0)? + 3-9) = /16 + 36 = V52
Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and
the perpendicular bisector of AB.
EXERCISE 7.1
1. Find the distance between the following pairs of points :
(i) (2,3), 4, D (ii) (-5,7), 1, 3) (iii) (a, b), (-a,-b)
2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance
between the two towns A and B discussed in Section 7.2. . "~~
3. Determine if the points (1, 5), (2, 3) and (-2, - 11) are collinear. _
4, Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
5. In a classroom, 4 friends are 10 :
seated at the points A, B, C and F | tT ] | tt fy
Das shown in Fig. 7.8. Champa 9 PT} fT | fd pd]
and Chameli walk into the class 8 =
and after observing for a few 7 | | { [ [plea | [|
minutes ChampaasksChamei, 7} {Tt | LTT tt
ors gow tink ABCP 8 # owe SFT ETT ET]
square?" Chameli disagrees. 5
Using distance formula, find . 4 | {| fal | [| [| ce
whichofem comet Cele
a 3
6 Nanetienpotamaitent 37 TPT TT
ormed, if any, by the following '
points, and give reasons for 1 tT TT | ty TT
your answer PET TT TT Tt tl
() C1,-2),(1,0),1,2,¢3,0 9 1 ? 3 _ . 67 8 9 10
.. olumns
(i) (-3, 5), 3, 1), (0, 3), (-l, -4)
(ii) (4,5), (7,6), 4,3), (1,2) Fig. 7.8
7, Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
8. Find the values of y for which the distance between the points P(2, - 3) and Q(10, y) is
10 units.
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106 MATHEMATICS
9, If Q(O, 1) is equidistant from P(5, -3) and R(, 6), find the values of x. Also find the
distances QR and PR.
10. Find a relation between x and y such that the point (x, y) is equidistant from the point
(3, 6) and (-3, 4).
7.3 Section Formula
Let us recall the situation in Section 7.2.
Suppose a telephone company wants to Y
position a relay tower at P between A and B
is such a way that the distance of the tower B(36, 15)
from B is twice its distance from A. If P lies (x, y) P
on AB, it will divide AB in the ratio | : 2 lL» Cc.
(see Fig. 7.9). If we take A as the origin O, A X
and 1 km as one unit on both the axis, the 0 Ds. E
coordinates of B will be (36, 15). In orderto - a
know the position of the tower, we must know _ Fig. 19
the coordinates of P. How do we find these _ ° "
coordinates? . oS °
Let the coordinates of P be (x, y).. Draw perpendiculars from P and B to the
x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by
the AA similarity criterion, studied in Chapter 6, APOD and A BPC are similar.
Theref OD _ OPS 1 geo oP}
ererer'> PC PB 2?" BC. PB 2
~ a s
- 1
So, _x _! and _y Lt,
36-x 2 Is-y 2 Y
These equations give x = 12 and y=5S. Bey ¥2)
You can check that P(12, 5) meets the m
condition that OP : PB = 1 : 2. OVP C
Now let us use the understanding that my 4
you may have developed through this A
example to obtain the general formula. (x )
p71
Consider any two points A(x,, y,) and
B(x,, y,) and assume that P (x, y) divides
AB internally in the ratio m, : m,, i.e., 0 R S T xX
PA m .
- = - (see Fig. 7.10). i
PB m, ( g ) Fig. 7.10
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CoorDINATE GEOMETRY 107
Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to
the x-axis. Then, by the AA similarity criterion,
A PAQ ~ A BPC
PA AQ PQ
Theref -= == 1
errr BP PC BC ()
Now, AQ =RS = OS - OR=x-x,
PC =ST=OT-OS=x,-x
PQ = PS-- QS = PS-- AR=y-y, |
BC = BT- CT = BT-PS=y,-y
Substituting these values in (1), we get .
Mm *7KH_ ITM -
My Xy-X YY - . _
m x-xXx . MX» + MND X,
Taking - = -- we get x= 4 -*+ .
My Xy- X . . m +m, _
m - me +
Similarly, taking et Fe get y = me FN
my y27y m, +m,
So, the coordinates of the point P(x, y) which divides the line segment joining the
points AG, yp) and BG,, Vo)s internally, in the ratio m, :m, are
(~ MX, + M4% , mh Y2 +m, y, (2)
SY im +m + m+m
This is known as the section formula.
This can also be derived by drawing perpendiculars from A, P and B on the
y-axis and proceeding as above.
If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be
. . key t+ 1 kyn +
k+1 k+1
Special Case : The mid-point of a line segment divides the line segment in the ratio
1: 1. Therefore, the coordinates of the mid-point P of the join of the points A(,, y,)
and B(x,, y,) is
I-x,+1-x, ly, +]l-y,)_ (4 +x. y+.
1+1 1+1 2 2 .
Let us solve a few examples based on the section formula.
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108 MATHEMATICS
Example 6 : Find the coordinates of the point which divides the line segment joining
the points (4, - 3) and (8, 5) in the ratio 3 : 1 internally.
Solution : Let P(x, y) be the required point. Using the section formula, we get
3(8) + 1(4) 3(5)+ 1(-3)
x= ----=7, y= -- 83
34+1 34+1
Therefore, (7, 3) is the required point.
Example 7 : In what ratio does the point (- 4, 6) divide the line segment joining the
points A(- 6, 10) and B(3, - 8)?
Solution : Let (- 4, 6) divide AB internally in the ratio m, : m,. Using the section
formula, we get .
3m,--6m, -8m, +10 .
ca,= [MES 2, ™) | ()
m+m, - mtM po
Recall that if (x, y) = (a, b) thenx=aandy=b. | " -
3my- 6m, . _, 8m, +10
So, 4-200" and 6 =m
m +m, m+ mM,
N 4 3m, - 6m, .
ow, a = m, + m, gives us
- 4m, - 4m, = 3m, - 6m,
1e., 7m, = 2m,
Le, m,:m,=2:7
You should verify that the ratio satisfies the y-coordinate also.
"em + 10 -8"1 +10
= + m
Now, a ee =, (Dividing throughout by m,)
m, +My), Mm +]
-8 x 2 +10
= -- =6
-+1
7
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CoorDINATE GEOMETRY 109
Therefore, the point (- 4, 6) divides the line segment joining the points A(- 6, 10) and
B(3, - 8) in the ratio 2 : 7.
Alternatively : The ratio m, : m, can also be written as ma :1, ork: 1. Let (4, 6)
My
divide AB internally in the ratio k : 1. Using the section formula, we get
C4,6)= 3k --6 -8k+10 (2)
kL ok +1 /
3k --6
So, -4= -
k+1 .
i.e., -4k--4= 3k-6 © :
i.e., Tk=2 ° ce
i.e., kil=2:7 o
You can check for the y-coordinate also. Soe "
So, the point (- 4, 6) divides the line segment joining the points A(- 6, 10) and
B(3, - 8) in the ratio 2: 7.
Note : You can also find this ratio by calculating the distances PA and PB and taking
their ratios provided you know that A, P and B are collinear.
pO Y -
Example 8 : Find the coordinates of the points of trisection (i.e., points dividing in
three equal parts) of the line segment joining the points A(2, - 2) and B(- 7, 4).
Solution ; Let P and Q be the points of A P Q B
trisection of AB i.e., AP = PQ = QB (2, -2) (-7, 4)
(see Fig. 7.11). Fig. 7.11
Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by
applying the section formula, are
\(-7) + 2(2) 14) + 2(-2) |
L T42) 142 debe C10)
Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
2(-7) +122) 2(4) + 1-2) | |
241 241. Shes 4,2)
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110 MATHEMATICS
Therefore, the coordinates of the points of trisection of the line segment joining A and
B are (-1, 0) and (- 4, 2).
Note : We could also have obtained Q by noting that it is the mid-point of PB. So, we
could have obtained its coordinates using the mid-point formula.
Example 9 : Find the ratio in which the y-axis divides the line segment joining the
points (5, - 6) and (-1, - 4). Also find the point of intersection.
Solution ; Let the ratio be k : 1. Then by the section formula, the coordinates of the
point which divides AB in the ratio k : 1 are Th+ 5 Ak 6),
k+1 k+l
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
Theref ktS gg a
erefore, ke] =. :
So, k=5 0
That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as
|
(o =) | oo.
3 a -
Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p.
Solution ; We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
. 7 6+9 144) (8+p 2+3
vee 2° 2) 2° 2
: 15 5 8+p 5
Le, - Bale l ta US
2 2 2 2
15 8+p
aa 27 2
1.¢., p=7
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CoorDINATE GEOMETRY 111
EXERCISE 7.2
1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the
ratio 2:3.
2. Find the coordinates of the points of trisection of the line segment joining (4, -1)
and (-2, -3).
3. To conduct Sports Day activities,in p Cc
your rectangular shaped school
ground ABCD, lines have been
drawn with chalk powder at a /
distance of Imeach. 100 flower pots ¥4
have been placed at a distance of 1m al
from each other along AD, as shown NG
LJ
1 .
in Fig. 7.12. Niharika runs a th the Ne .
WV .
tl :
distance AD on the 2nd line and us i; *
1 WG I; : : jj Me
posts a green flag. Preet runs 5 th a oo fe
the distance AD on the eighth line v - ,
and posts a red flag. What is the -~ Uy 2
distance between both the flags? If ¥7 1
Rashmi has to post a blue flag exactly A B
halfway between the line segment. 123 45 678 9 10
joining the two flags, where should :
she post her flag? ) Fig. 7.12
4. Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided
by © 1,6).
5. Find the ratio in which the line segment joining A(1, - 5) and B(- 4, 5) is divided by the
x-axis. Also find the coordinates of the point of division.
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
xandy. ~~
7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is
(2,-- 3) and B is (1, 4).
8. If A and B are (- 2, -2) and (2, - 4), respectively, find the coordinates of P such that
3
AP= 7 AB and P lies on the line segment AB.
9. Find the coordinates of the points which divide the line segment joining A(- 2, 2) and
B(2, 8) into four equal parts.
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (- 1, 4) and (2, - 1) taken in
1
order. [Hint : Area of a rhombus = 5 (product of its diagonals)]
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112 MATHEMATICS
7.4 Summary
In this chapter, you have studied the following points :
1. The distance between P(x,, y,) and Q(x,, y,) is J, - 4) +O. - y,)'.
2. The distance of a point P(x, y) from the origin is x? + y'.
3. The coordinates of the point P(@, y) which divides the line segment joining the
points A(x,, y,) and B(x,, y,) internally in the ratio m, : m, are
MyXy + MX MY. +MY Y,
m, + m, m, +m, ,
4. The mid-point of the line segment joining the points P(x,, y,) and Q(x,, y,) is
[Se Ase) .
2 2 ) ;
A NOTETO THE READER
Section 7.3 discusses the Section Formula for the coordinates (x, y) of a
point P which divides internally the line segment joining the points
A(x,, y,) and B(x,, y,) in the ratio m, : m, as follows :
MX, + MX ~_ My, +My,
x= > ys
. m+m m +m,
Note that, here, PA: PB =m, : m,.
However, if P does not lie between A and B but lies on the line AB,
outside the line segment AB, and PA: PB =m, : m,, we say that P divides
externally the line segment joining the points A and B. You will study
Section Formula for such case in higher classes.
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[a] es Le]
7 a
(a) ee
1062CHO8
There is perhaps nothing which so occupies the i
middle position of mathematics as trigonometry.
- J.F. Herbart (1890)
8.1 Introduction . . eT
You have already studied about triangles, and in particular, right triangles, in your
earlier classes. Let us take some examples from our surroundings where right triangles
can be imagined to be formed. For instance : , f
1. Suppose the students of a school are , a
visiting Qutub Minar. Now, if a student 4 we i
is looking at the top of the Minar, a right a E.
triangle can be imagined to be made, © wn -
as shown in Fig 8.1. Can the student we ;
find out the height of the Minar, without aw | :
actually measuringit? (ore If
2. Suppose a girl is sitting on the balcony verre
of her house located on the bank of a Fig. 8.1
river. She is looking down at a flower
pot placed ona stairofatemple situated
nearby on the other bank of the river. f-5.
A right triangle is imagined to be made > f Pa KA
in this situation as shown in Fig.8.2. If -_Na i me awa
you know the height at which the © 7] ma Arif
person is sitting, can you find the width os seas leet |
TE |
iver?
of the river? Fig. 8.2
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114 MATHEMATICS
3. Suppose a hot air balloon is flying in

the air. A girl happens to spot the

balloon in the sky and runs to her @, @)

mother to tell her about it. Her mother a at

rushes out of the house to look at the o weg

balloon.Now when the girl had spotted Obes

the balloon intially it was at point A. oC ae

When both the mother and daughter fe

came out to see it, it had already Ags A

travelled to another point B. Can you . '

find the altitude of B from the ground? Fig. 8.3

In all the situations given above, the distances or heights can be found by using
some mathematical techniques, which come under a branch of mathematics called
'trigonometry'. The word 'trigonometry' is derived from the Greek words 'tri'
(meaning three), 'gon' (meaning sides) and ."metron' (meaning measure). In fact,
trigonometry is the study of relationships between the sides and angles of a triangle.
The earliest known work on trigonometry was recorded in Egypt and Babylon. Early
astronomers used it to find out the distances of the stars and planets from the Earth.
Even today, most of the technologically advanced methods used in Engineering and
Physical Sciences are based on trigonometrical concepts.

In this chapter, we will study some ratios of the sides of a right triangle with
respect to its acute angles, called trigonometric ratios of the angle. We will restrict
our discussion to acute angles only. However, these ratios can be extended to other
angles also. We will also define the trigonometric ratios for angles of measure 0° and
90°. We will calculate trigonometric ratios for some specific angles and establish
some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Ratios Cc
In Section 8.1, you have seen some right triangles <
imagined to be formed in different situations. z

Let us take a right triangle ABC as shown Hypotenuse £
in Fig. 8.4, =

Here, Z CAB (or, in brief, angle A) is an 5
acute angle. Note the position of the side BC
with respect to angle A. It faces Z A. We call it =
the side opposite to angle A. AC is the A Side adjacent to B
hypotenuse of the right triangle and the side AB angle A
is a part of Z A. So, we call it the side .
adjacent to angle A. Fig. 8.4

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INTRODUCTION TO TRIGONOMETRY 115
Note that the position of sides change Cc
when you consider angle C in place of A -
(see Fig. 8.5). <
You have studied the concept of 'ratio' in Ht 3
: . . ypotenuse &
your earlier classes. We now define certain ratios 3 Y
involving the sides of a right triangle, and call xz Sb
them trigonometric ratios. wa
The trigonometric ratios of the angle A A B
in right triangle ABC (see Fig. 8.4) are defined Side opposite to -
as follows : angle C
. side opposite to angle A | BC Fig. 8.5
sine of Z A = = -
hypotenuse AC
. side adjacent to angle A_- AB oo .
cosine of Z A= --W___--_ = -- :
hypotenuse _ AC _ -
side opposite to angle A BCtangent of Z A = ---_________ = -_
side adjacent to angle A = AB
1 hypot AC
cosecant of Z A= poise
sine of ZA side opposite to angle A BC
"oN 1 oh A
secant of ZA= - 1 ___- hypotenuse _ AC
'cosine of ZA side adjacent to angle A AB
4 ide adjacent to angle A AB
cotangent of Z A= -----_ = = em eo mee
tangent of Z A side opposite to angle ABC
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A
and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively,
the reciprocals of the ratios sin A, cos A and tan A.
BC
Also, observe that tan A= P= = AC = MA ag cota = SA
so, observe that tan A= 7 'AB cos A and cot A= sinA
AC
So, the trigonometric ratios of an acute angle in a right triangle express the
relationship between the angle and the length of its sides.
Why don't you try to define the trigonometric ratios for angle C in the right
triangle? (See Fig. 8.5)
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116 MATHEMATICS
The first use of the idea of 'sine' in the way we use
it today was in the work Aryabhatiyam by Aryabhata, he (Ray
in A.D. 500. Aryabhata used the word ardha-jya oH) SN
for the half-chord, which was shortened to jya or 4 ta
jiva in due course. When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as f 7
itis. The word jiva was translated into sinus, which j
means curve, when the Arabic version was translated
into Latin. Soon the word sinus, also used as sine, -
became common in mathematical texts throughout UJ
Europe. An English Professor of astronomy Edmund
Gunter (1581-1626), first used the abbreviated Aryabhata
notation 'sin'. C.E. 476 - 550
The origin of the terms 'cosine' and 'tangent' was much later. The cosine function
arose from the need to compute the sine of the complementary angle. Aryabhatta
called it kotijya. The name coesinus originated with Edmund Gunter. In 1674, the
English Mathematician Sir Jonas Moore first used the abbreviated notation 'cos'.
Remark : Note that the symbol sin A is used as an
abbreviation for 'the sine of the angle A'. sin A is not Q
the product of 'sin' and A. 'sin' separated from A c "
has no meaning. Similarly, cos A is not the product of .
'cos' and A. Similar interpretations follow for other P :
trigonometric ratios also. 7 .

Now, if we take a point P on the hypotenuse Hypotenuse :
AC or a point Q on AC extended, of the right triangle
ABC and draw PM perpendicular to AB and QN A ot
perpendicular to AB extended (see Fig. 8.6), how M BN
will the trigonometric ratios of Z A in A PAM differ
from those of Z Ain A CAB or from those of Z Ain Fig. 8.6
A QAN?

To answer this, first look at these triangles. Is A PAM similar to A CAB? From
Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the
triangles PAM and CAB are similar. Therefore, by the property of similar triangles,
the corresponding sides of the triangles are proportional.

AM AP _ MP
So, we have AB = AC BC
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INTRODUCTION TO TRIGONOMETRY 117
F hi find MP = BE- sin A
rom this, we fin AP = AC .
. AM AB MP BC
Similarly, -- = - = cos A, -- =- =tan A and so on.
AP AC AM AB
This shows that the trigonometric ratios of angle A in A PAM not differ from
those of angle A in A CAB.
In the same way, you should check that the value of sin A (and also of other
trigonometric ratios) remains the same in A QAN also. |
From our observations, it is now clear that the values of the trigonometric
ratios of an angle do not vary with the lengths of the sides of the triangle, if
the angle remains the same. .
Note : For the sake of convenience, we may write sin'A, cos?A, etc., in place of
(sin A)', (cos A)', etc., respectively. But cosec A = (sin A)! ¥ sin"! A (it is called sine
inverse A). sin"! A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well. Sometimes, the
Greek letter 6 (theta) is also used to denote an angle. .
We have defined six trigonometric ratios of an acute angle. If we know any one
of the ratios, can we obtain the other ratios? Let us see.
1 , Cc
If in a right triangle ABC, sin A= 3° .
then thi th cp Bell i.e., th 3 k
en this means that 7 (=, ie., the .
lengths of the sides BC and AC of the triangle
ABC are in the ratio 1 : 3 (see Fig. 8.7). Soif A B
BC is equal to k, then AC will be 3k, where Fig. 8.7
kis any positive number. To determine other
trigonometric ratios for the angle A, we need to find the length of the third side
AB. Do you remember the Pythagoras theorem? Let us use it to determine the
required length AB. ,
AB? = AC? - BC? = (3k)? - (KY = 82 = (2V2 ky
Therefore, . AB= + 2/2 k
So, we get AB= 2V2k (Why is AB not --2V2k2)
x ae ABE 22k _ 2/2
om oS AC 3k
Similarly, you can obtain the other trigonometric ratios of the angle A.
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118 MATHEMATICS
Remark : Since the hypotenuse is the longest side in a right triangle, the value of
sin A or cos A is always less than 1 (or, in particular, equal to 1).
Let us consider some examples.
4 Cc
Example 1 : Given tan A = 3° find the other
trigonometric ratios of the angle A.
Solution : Let us first draw a right A ABC 4k
(see Fig 8.8).
N k h A Bef ,
ow, we know that tan A= AB 3°
Therefore, if BC = 4k, then AB = 3k, where k is a A 3k B
positive number. . ,
j ~ Fig. 8.8
Now, by using the Pythagoras Theorem, we have OO
AC' = AB* + BC? = (4k) + (3k)? = 25k?
So, AC=5k . sO
Now, we can write all the trigonometric ratios using their definitions.
mae BOM 4
me=AC 5k 5
y= AB 3k 3
EON AC SKS
Th f tA _1 23 cosee Az lL => d A I 5
ePETOREs COLIN tanA 4 sinA 4 anesee A= cosA 3
Los P
Example 2: If Z B and Z Q are
acute angles such that sin B = sin Q, A
then prove that Z B = Z Q.
Solution : Let us consider two right
triangles ABC and PQR where
ae .; Cc BOR Q
sin B = sin Q (see Fig. 8.9).
Fig. 8.9
. AC
We have sin B= AB
; now Ee
an sin Q= PQ
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INTRODUCTION TO TRIGONOMETRY 119
th AC _ PR
n AB PQ
Theref AC, MB Lk sa 1
erefore, PR = PQ." y (1)
Now, using Pythagoras theorem,
BC = J AB?- AC'
and QR = PQ' - PR? :
BC VAB?-AC? /k?PQ?-k?PR? -_k./PQ?- PR? k am
So, Ay = Rw o.oo ee 2
* QR ~ /pg?-PR? -,/PQ?- PR? «PQ? - PR?» | 2)
From (1) and (2), we have Ce
AC _AB_BC
PR PQ. QR ~ .
Then, by using Theorem 6.4, A ACB ~ A PRQ and therefore, Z B= ZQ.
A
Example 3 : Consider A ACB, right-angled at C, in
which AB = 29 units, BC = 21 units and Z ABC =8
(see Fig. 8.10). Determine the values of |
{ ' ° 29
(i) cos? 6 + sin' 6, \_ ' .
(ii) cos? 6 - sin? 0. :
Solution : In A ACB, we have c (\N B
21
_- 2 2 2 2
AC = AB -- BC = ¥(29)° - (21) Fig. 8.10
= f(29 - 21)(29 + 21) = ,/(8)(50) = V400 = 20units
S Sine="AB 29°C ABO
Now. () cos?@ + sina = {20 * (21) _ 20+ 27 _ 400+ 441 _,
ow, (i) cos' + sin'@ = 9 29 992 ga)
Gi) cos? - sin? = | 2 * (20) _ (21+ 20)(21-20) 41
and (ii) cos* 9 - sin' 6 = 29 29 992 841°
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120 MATHEMATICS
Example 4: In a right triangle ABC, right-angled at B, A
if tan A = 1, then verify that
2 sinAcosA= 1.
BC .
Solution : In A ABC, tan A= AB = 1 (see Fig 8.11)
B Cc
Le., BC = AB
Fig. 8.11
Let AB = BC = k, where k is a positive number.
Now, AC = ./AB'+ BC? i
= (k)? + (k)? =k V2. |
ae Beet AB
Therefore, sin A= AC V2. and cosA= AC 2
So 2 sin A cos A= (+5) (=| = 1 which is the required value
V2)\V2)
Example 5 : In A OPQ, right-angled at P, Q
OP = 7 cm and OQ - PQ = 1 cm (see Fig. 8.12).
Determine the values of sin Q and cos Q.
Solution : In A OPQ, we have ~
< OQ? = OP? + PQ?
Le., "ad + PQ) = OP? + PQ? (Why?)
Le., 1 + PQ? + 2PQ = OP? + PQ?
ie, _ 14+2PQ= 7 (Why?)
P 7 oO
i.e., PQ = 24cm and OQ =1+4+PQ=25cm em
Fig. 8.12
5 | ond 24
0, sinQ= 5 an cos Q= 55
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INTRODUCTION TO TRIGONOMETRY 121
EXERCISE 8.1
1. InAABC, right-angled at B, AB = 24cm, BC =7 cm. Determine :
(i) sin A, cos A P
(i) sin C,cosC
2. In Fig. 8.13, find tan P-cotR.
3 12 cm 13 cm
3. Ifsin A= ra calculate cos A and tan A.
4, Given 15 cot A = 8, find sin A and sec A.
13 8 R
5. Given sec 0= DD' calculate all other trigonometric ratios. Fig. 8.13
6. If Z A and Z B are acute angles such that cos A = cos B, then show that Z A = ZB.
7 1+ sin 6)(1 - sin 0 .
7. IfcotO= -> evaluate: (i) (1+ sin 6)(1= sin 6), (ii) cot? @
8 (1 + cos 8)(1 - cos) oo ;
1- tan? A , : |
8. If 3 cot A=4, check whether 7--__73,_ =cos? A - sin?A or not.
1+tan'A -
1
9. In triangle ABC, right-angled at B, if tan A= BR find the value of:
(i) sin A cos C+cos A sin C
(i) cos A cos C- sin A sinC a
fr
10. In APQR, right- angled at Q, PR + QR = 25 cm and PQ =5 cm. Determine the values of
sin P, cos P and tan P." .
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
12 >
(i) sec A= 5 for some value of angle A.
(itt) cos A is the abbreviation used for the cosecant of angle A.
(tv) cot A is the product of cot and A.
. 4
(v) sin6= 3 for some angle 0.
8.3 Trigonometric Ratios of Some Specific Angles
From geometry, you are already familiar with the construction of angles of 30°, 45°,
60° and 90°. In this section, we will find the values of the trigonometric ratios for these
angles and, of course, for 0°.
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122 MATHEMATICS
Trigonometric Ratios of 45° C
In A ABC, right-angled at B, if one angle is 45°, then
the other angle is also 45°, ie., Z A= ZC = 45°
(see Fig. 8.14).
So, BC =AB (Why?)
A B
Now, Suppose BC = AB = a. Fig. 8.14
Then by Pythagoras Theorem, AC" = AB? + BC? = a? + a? = 2a',
and, therefore, AC = ayJ2- :
Using the definitions of the trigonometric ratios, we have :
in 45° side opposite to angle 45° BC a 1. ,
in = -. Sa FF Cr
° hypotenuse AC a2 2: -
45° = side adjacent toangle 45° AB a | 1.
cos = hypotenuse . AC aV2 2
side opposite to angle 45° BC a
tan 45° = = - = - = 1
side adjacent to angle 45° AB a
In od I
Also, cosec 45° = -- = v2, sec 45° = -- =y2, cot 45° = --- =].
sin 45 y cos 45 tan 45°
SS .
Trigonometric Ratios of 30° and 60° A
Let us now calculate the trigonometric ratios of 30° [|
and 60°. Consider an equilateral triangle ABC. Since
each angle in an equilateral triangle is 60°, therefore, 30°
ZA=ZB=2ZC=60°.
Draw the perpendicular AD from A to the side BC B A60° Cc
(see Fig. 8.15). _?P
Now A ABD = AACD (Why?) Fig. 8.15
Therefore, - BD= DC
and ZBAD= Z CAD (CPCT)
Now observe that:
AABD is aright triangle, right- angled at D with Z BAD = 30° and Z ABD = 60°
(see Fig. 8.15).
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INTRODUCTION TO TRIGONOMETRY 123
As you know, for finding the trigonometric ratios, we need to know the lengths of the
sides of the triangle. So, let us suppose that AB = 2a.
1
Then, BD= 5 BC =a
and AD? = AB? - BD? = (2a)? - (a)' = 3a',
Therefore, AD= av3
Now, we have : .
130° eee Had coe 392 = AD 3 _ 93
sin 30°= -- =-=-, cos = = =
AB 2a 2 AB 2a 2
o_ BPL a bt 4
tan 30° = AD a3 3° | .
Al sore =, cee3q22 a
so, cosec = Gn30 sec = cos 30° 3 :
cot 30° = ---= 3. oe -
tan 30° . .
Similarly, ;
moor= AD - 23 53 oo. 692 = 4, tan 60° B
sin 60° = -- = -- = -., cos = -, tan = /3,
AB 2a 2°.» 2 3
60° <= see 60° = 2 and cot 60° =
cosec = 3) sec = 2 and cot = 3B
ML .
Trigonometric Ratios of 0° and 90°
Let us see what happens to the trigonometric ratios of angle C
A, if itis made smaller and smaller in the right triangle ABC
(see Fig. 8.16), till it becomes zero. As Z A gets smaller and
smaller, the length of the side BC decreases.The point C gets
closer to point B, and finally when Z A becomes very close A ; B
to 0°, AC becomes almost the same as AB (see Fig. 8.17). Fig. 8.16
Cc.
" ot A of 4
° Cc oe . ' oo 7
Y " AC "ot "od wot
4 ° a . iC
A BA B A B A B A B A B
Fig. 8.17
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124 MATHEMATICS
When Z A is very close to 0°, BC gets very close to 0 and so the value of
BC
sinA= AC is very close to 0. Also, when Z A is very close to 0°, AC is nearly the
AB
same as AB and so the value of cos A= AC is very close to 1.
This helps us to see how we can define the values of sin A and cos A when
A = 0°. We define : sin 0° = 0 and cos 0° = 1.
Using these, we have : °
sin 0° 1 ar .
tan 0° = --- =0, cot 0° = --->: which is not defined. (Why?)
cos 0° tan 0° .
1 ra ae
sec 0° = --- =] and cosec 0° = ----; which is again not defined.(Why?)
cos 0° sin 0° eo
Now, let us see what happens to the trigonometric ratios of Z A, when it is made
larger and larger in A ABC till it becomes 90°. As Z A gets larger and larger, Z C gets
smaller and smaller. Therefore, as in the case above, the length of the side AB goes on
decreasing. The point A gets closer to point B. Finally when Z A is very close to 90°,
Z C becomes very close to 0° and the side AC almost coincides with side BC
(see Fig. 8.18). a :
Cc Cle ¢ C c
Li] J : A, ol] od
A B a8 A a B ae
, . Fig. 8.18
When Z C is very close to 0°, Z A is very close to 90°, side AC is nearly the
same as side BC, and so sin A is very close to 1. Also when Z A is very close to 90°,
Z Cis very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.
So, we define : sin 90° = 1 and cos 90° = 0.
Now, why don't you find the other trigonometric ratios of 90°?
We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60°
and 90° in Table 8.1, for ready reference.
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INTRODUCTION TO TRIGONOMETRY 125
Table 8.1
za | oe | so | ase | oo |
; 1 1 v3
sinA 0 7 V2 > 1
V3 1 1
cos A 1 > V2 7 0
A 0 4 1 Not defined
tan B V3 ot define
A | Not defined 2 = 1
cosec ot define V2 B
A 1 = 2 Not defined
sec B J2 ot define
A Not defined 1 + 0
cot ot define V3 B
Remark : From the table above you can observe that as Z A increases from 0° to
90°, sin A increases from 0 to 1 and cos A decreases from | to 0.
/
Let us illustrate the use of the values in the table above through some examples.
Example 6 : In A ABC, right-angled at B, A
AB = 5 cm and Z ACB = 30° (see Fig. 8.19).
Determine the lengths of the sides BC and AC.
. 5
Solution : To find the length of the side BC, we will"
choose the trigonometric ratio involving BC and the -~
given side AB. Since BC is the side adjacent to angle B c
C and AB is the side opposite to angle C, therefore Fig. 8.19
- AB anc
BC tan
+ 5 30° a
1.¢., BC tan = 3
which gives BC= 5/3 cm
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126 MATHEMATICS
To find the length of the side AC, we consider

in 30° AB Why?

sin = AC (Why?)
; 1S
1.¢., 2 = AC
i.e., AC = 10cm
Note that alternatively we could have used Pythagoras theorem to determine the third
side in the example above,
ice., AC = JAB? + BC? = ,/5?+ (5V3)? cm = 10cm.
Example 7 : In A PQR, right-angled at ~ So
Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. * PrP.
Determine Z QPR and Z PRQ. - -
Solution : Given PQ=3cmandPR=6cm.. cm 6cm

PQ.
Therefore, PR = sinR Q R

Fig. 8.20

nredel
or oye 62
So, "ZPRQ = 30°
and therefore, Z QPR = 60°. (Why?)
You may note that if one of the sides and any other part (either an acute angle or any
side) of a right triangle is known, the remaining sides and angles of the triangle can be
determined. no

Ss 1 1
Example 8 : If sin (A- B) = 7° 008 (A+ B)= 2" 0° <A+B< 90°, A> B, find A
andB. . |
: 1

Solution : Since, sin (A - B) = ~~ therefore, A- B = 30° (Why?) (1)

1
Also, since cos (A + B) = 2? therefore, A+ B =60° (Why?) (2)
Solving (1) and (2), we get: A= 45° and B = 15°.

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INTRODUCTION TO TRIGONOMETRY 127
EXERCISE 8.2
1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan? 45° + cos? 30° - sin? 60°
(iti cos 45° (iv) sin 30° + tan 45° - cosec 60°
ii 3 3 wv a
sec 30° + cosec 30 sec 30° + cos 60° + cot 45°
5 cos" 60° + 4 sec" 30° - tan" 45° '
(v) +2 fo} 2 fe}
sin* 30° + cos" 30
2. Choose the correct option and justify your choice : a
@ 2 tan 30° . . ,
i) -,-= ",
1+ tan? 30° . on
(A) sin60° (B) cos 60° . (©) tan 60° _* © sin30°
_ I- tan" 45° _ . °
(i) 1+ tan? 45° . ,
(A) tan 90° ®) 1 © sin45° @) 0
(iii) sin2A=2 sin Ais true whenA=. ,
(A) o@ f° ) (B) 30° - O48 (D) 60°
NL .
2 tan 30° :
() T= tan? 30°
(A) cos 60°, (B) sin 60° (C) tan 60° (D) sin 30°
a , 1
3. Iftan(A+B)= ./3 and tan(A--B)= B®" <A+B<90°; A>B, find A and B.
4. State whether the following are true or false. Justify your answer.
@) sin(A+B)=sinA+sinB.
(ii) The value of sin 6 increases as 0 increases.
(iit) The value of cos 6 increases as 9 increases.
(iv) sin @=cos 9 for all values of 6.
(v) cot Ais not defined for A= 0°.
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128 MATHEMATICS
8.4 Trigonometric Identities A
You may recall that an equation is called an identity
when it is true for all values of the variables involved.
Similarly, an equation involving trigonometric ratios
of an angle is called a trigonometric identity, if it is
true for all values of the angle(s) involved.
In this section, we will prove one trigonometric
identity, and use it further to prove other useful C , B
trigonometric identities. Fig. 8.21
In A ABC, right-angled at B (see Fig. 8.21), we have:
AB? + BC? = AC? © dd)
Dividing each term of (1) by AC', we get | .
AB? BC? AC? a |
- + - = - ;
AC? AC? ACH .
ABY (BC) (ACY -
1.¢., Sn] Tian] HR lazn -
AC AC LAC
1.¢., (cos A) + (sin A)? = 1 :
Le., cos? A + sin? A = 1° ° (2)
This is true for all A such that 0° < A < 90°. So, this is a trigonometric identity.
Let us now divide (1) by AB. We get
AB' BC? AC"
--_ + -_- = ~--
_ , AB? AB* - AB?
_. {ABY (2c) (Ac)
Or, . ssi tlao] ERE la
AB AB AB
ie., 1+ tan'A = sec? A (3)
Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and
sec A are not defined for A = 90°. So, (3) is true for all A such that 0° < A < 90°.
Let us see what we get on dividing (1) by BC". We get
AB' BC? AC?
Re) Re = pe
BC' BC BC
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INTRODUCTION TO TRIGONOMETRY 129
AB) (BC) (Acy
1.¢., sa] tion] =lan
BC BC BC
i.e., cot? A + 1 = cosec? A (4)

Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for
all A such that 0° < A< 90°,

Using these identities, we can express each trigonometric ratio in terms of other
trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the
values of other trigonometric ratios.

Let us see how we can do this using these identities. Suppose we know that

] .
tan A= B Then, cot A= /3. . . -
2° V3
Since, sec? A= 1+ tan?A= 1414, sec A= -=, andcosA= V3.
393 0°. VB 2
Again, sin A= /1-cos?A =,/1- , = = Therefore, cosec A = 2.
Example 9 : Express the ratios cos A, tan A and sec A in terms of sin A.
fo N
Solution: Since. cos? A + sin' A= 1, therefore,
cos? A= 1-sin? A, ie, cos A= +1 -sin?A
This gives i cosA= ,/l-sin? A (Why?)
H aan SA sin A and sec A 1 1
ence, taa A = -- = =
cosA J/1-sin'?A cos A Jl-sin? A
Example 10 : Prove that sec A (1 - sin A)(sec A+ tan A) = 1.
Solution :
LHS A (1 -sin A)(sec A A 1 (1 - sin A) _1_,snA
= sec A(1 - sin A)(sec A + tan A) = cos A cos A cosA
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130 MATHEMATICS
(l1--sin A)(1+sin A) 1-- sin? A
~ cos' A cos' A
cos' A | 1 RH
~ cos? A = RHS
tA- A A-1l
Example 11 : Prove that aoe ore
cotA+cosA cosecA+l1
LA A OS - cos A :
Solution : LHS = SA =c0s A _ sim A |
cot A+ cos A +cos A -
sin A .
cos Af 1] [o-1] . A 4
sim sin " _
_ ; _ = cosec - = RHS
+
cos A| --+1 - +1 cose
snA .} _\snA
in @-cos 6 +1 1 ,
Example 12 : Prove that "ee = --_---> using the identity
sin6+cos@-1 sec @-tan@
sec? 6 = | + tan' 6. | - .
Solution : Since we will apply the identity involving sec 9 and tan 9, let us first
convert the LHS (of the identity we need to prove) in terms of sec 6 and tan @ by
dividing numerator and denominator by cos 0.
sin@-cos@+1 tan@-1+sec@
LHS = +. =.
sin6@+cos@-1 tan@+1-sec@
_ (tan @+secO)-1 _ {(tan 0+ sec 9) - 1} (tan 6 - sec 0)
~ (tan @-sec@)+1 {(tan 6 - sec 0) + 1} (tan @- sec 6)
: (tan? @ - sec" ) - (tan 6 - sec 0)
= {tan 8 - sec 8 + 1} (tan®@ - sec 8)
_ -1- tan 6+ sec 8
~ (tan 0 - sec 6+ 1) (tan 6 - sec 6)
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INTRODUCTION TO TRIGONOMETRY 131
-1 _ 1
~ tan@-sec@ sec @- tan @
which is the RHS of the identity, we are required to prove.
EXERCISE 8.3
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2. Write all the other trigonometric ratios of Z Ain terms of sec A. .
3. Choose the correct option. Justify your choice.
@) 9sec?A-9 tan? A= ,
(A) 1 (B) 9 © 8 Oo |
(ii) (1 + tan 6 + sec 9) (1 + cot 8 -cosec 8) = - . .
(A) 0 (B) 1 ~ ©2 oe (D) -1
(ii) (secA+tanA)(1-sinA)= a -
(A) sec A (B) sinA ~ ©) cosec A : (D) cos A
-- L+tan? A | ; :
(iv) 1+cot? A
(A) sec? A ®) -1 - ©}cot?A (D) tan? A
4. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
Ne x
@ ( 6 10)? 1- cos 8 (i) cos A , Ltsin A > sec A
i) (cosec 8-cot 6)? = ii) ----_ + ---- =
1 +cos 0 1+sinA cos A
(iii) _tan 6 + cot 8 =1+ sec 6 cosec 9
1-cot®@ 1-tan9
[Hint : Write the expression in terms of sin 9 and cos 8]
. l+sec A _ sin? A . oo.
(iv) secA L-cosA {Hint : Simplify LHS and RHS separately]
(v) cos A = sin A+) _ cosec A + cot A, using the identity cosec? A= 1+ cot? A.
cos A + sin A -1
1+sinA in 9-2 sin?
(vi) avs =secA+tanA (vii) Sinyaesin o 2 sin' ® = tan 0
V1 -sin A 2 cos" 6 - cos 6
(viii) (sin A + cosec A)'+ (cos A+ sec A)? =7 + tan?A+cot?A
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132 MATHEMATICS
x) (cosec A - sin A)(sec A - cos A) = a
(ix) tanA + cot A
[Hint : Simplify LHS and RHS separately]
2 2
[+tan°A | {l-tanA ;
@) 1+ cot?A 1-cot A} ~ ar A
8.5 Summary
In this chapter, you have studied the following points : .
1. Inaright triangle ABC, right-angled at B,
sin A= side opposite to angle AY cos A = side adjacent to angle A
hypotenuse hypotenuse .
side opposite toangle A
tan A= .
side adjacent to angle A "
2. cosee A = ---: sec A= -/_: tan A= -1-> tan A= SA :
sin A cosA cotA . cos A
3. Ifone of the trigonometric ratios of an acute angle is known, the remaining trigonometric
ratios of the angle can be easily determined. .
4, The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.
5. The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° < A< 90°) or
cosec A (0° <A < 90°) is always greater than or equal to 1.
in2 2 = - :
6. sin? A+cos?A=1, ~.
sec? A-tan*A=1 for.0°<A<90°, °
cose'? A=1+cot?A for 0°<A<90°
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(a) ees
nade hing
PRA: oe
I i
foes =
1062CHO9
9.1 Heights and Distances . :
In the previous chapter, you have studied about trigonometric ratios. In this chapter,
you will be studying about some ways in which trigonometry is used in the life around
you. oo ' "
Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1.
we) = a
X o a hye ' . ' a :
ro . : mY * .
yee ; I yee :
" " ;
ae Anseetdteratin i; ao Anse ofeeration ad
Fig. 9.1
In this figure, the line AC drawn from the eye of the student to the top of the
minar is called the line of sight. The student is looking at the top of the minar. The
angle BAC, so formed by the line of sight with the horizontal, is called the angle of
elevation of the top of the minar from the eye of the student.
Thus, the line of sight is the line drawn from the eye of an observer to the point
in the object viewed by the observer. The angle of elevation of the point viewed is
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134 MATHEMATICS
the angle formed by the line of sight with the horizontal when the point being viewed is
above the horizontal level, i.e., the case when we raise our head to look at the object
(see Fig. 9.2),
Object
g
ay
be got
© ot
yw CPi
peel. ae
Foro hoses ces ececc + Horizontal level "
o)
re :
Fig. 9.2 .

Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is
looking down at a flower pot placed on a stair of the temple. In this case, the line of
sight is below the horizontal level. The angle so formed by the line of sight with the
horizontal is called the angle of depression. , ° -

Thus, the angle of depression of a point on the object being viewed is the angle
formed by the line of sight with the horizontal when the point is below the horizontal
level, i.c., the case when we lower our head to look at the point being viewed
(see Fig. 9,3), ae

- : ' .
ee
d Angle of
A das depress;
a a rh Pression
| , (eo) Object
Fig. 9.3

Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3.
Are they angles of elevation or angles of depression?

Let us refer to Fig. 9.1 again. If you want to find the height CD of the minar
without actually measuring it, what information do you need? You would need to know
the following:

(i) the distance DE at which the student is standing from the foot of the minar
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SOME APPLICATIONS OF TRIGONOMETRY 135
(ii) the angle of elevation, Z BAC, of the top of the minar
(iii) the height AE of the student.

Assuming that the above three conditions are known, how can we determine the
height of the minar?

In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student.

To find BC, we will use trigonometric ratios of Z BAC or Z A.

In A ABC, the side BC is the opposite side in relation to the known Z A. Now,
which of the trigonometric ratios can we use? Which one of them has the two values
that we have and the one we need to determine? Our search narrows down to using
either tan A or cot A, as these ratios involve AB and BC.
Theref as?e tase hich Ivi ld gi BC

erefore, tanA= Ap ° COtA= GG' Which on solving woul give us BC.
By adding AE to BC, you will get the height of the minar. - we

Now let us explain the process, we have just discussed, by solving some problems.
Example 1 : A tower stands vertically on the ground. From a point on the ground,
which is 15 m away from the foot of the tower, the angle of elevation of the top of the
tower is found to be 60°. Find the height of the tower.

Solution : First let us draw a simple diagram to
represent the problem (see Fig. 9.4). Here AB d
represents the tower, CB is the distance of the point fi
from the tower and Z ACB is the angle of elevation. es
We need to determine the height of the tower, i.c., xy
AB. Also, ACB is a triangle, right- angled at B. AN
To solve the problem, we choose the trigonometric A
ratio tan 60° (or cot 60°), as the ratio involves AB Wi
mo ve Lid
and BC. ' \
N tan 60° A
ow, an = Be WYN
" cO =
AB <- 15m->B
ie. 3 =
15 Fig. 9.4
ie. AB = 15V3
Hence, the height of the tower is 15/3 m.
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136 MATHEMATICS
Example 2 : An electrician has to repair an A
electric fault on a pole of height m. She needs | === {== Fto reach a point 1.3m below the top of the pole -*
to undertake the repair work (see Fig. 9.5). What B™
should be the length of the ladder that she should ' S
use which, when inclined at an angle of 60° to
the horizontal, would enable her to reach the
required position? Also, how far from the foot e
of the pole should she place the foot of the a:
ladder? (You may take ./3 = 1.73) et .
Solution : In Fig. 9.5, the electrician is required to ne °
reach the point B on the pole AD. t | _ 60
. D \ Cc
So, BD =AD - AB = (5 - 1.3)m =3.7 m. . Fig. 9.5
Here, BC represents the ladder. We need to find. FC ,
its length, i.e., the hypotenuse of the right triangle BDC. _
Now, can you think which trigonometic ratio should we consider?
It should be sin 60°. | oe
BD 3.7 - V3

So, BC" sin 60° o Bc 4

cy .
Theref \ BC STK? _ 4.08 ( )

erefore, = ->=- =4.28 m (approx.
OB PP
1.¢e., the length of the ladder should be 4.28 m.
N " DC t 60° a
ow, _. BDS co "Bh
1 | DC 37 2.14
Le, | | = BR =? m (approx.)
Therefore, she should place the foot of the ladder at a distance of 2.14 m from
the pole.
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SOME APPLICATIONS OF TRIGONOMETRY 137
Example 3: An observer 1.5 m tall is 28.5 m away A
from a chimney. The angle of elevation of the top tn
of the chimney from her eyes is 45°. What is the uIL
height of the chimney? TIT
|
Solution : Here, AB is the chimney, CD the veer
observer and Z ADE the angle of elevation (see oll
Fig. 9.6). In this case, ADE is a triangle, right-angled 45°. tne, | Hea E
at E and we are required to find the height of the D rfl
chimney. f a
Cc B

We have AB= AE+BE=AE+1.5 Fig. 9.6 7
and DE= CB =28.5m "

To determine AE, we choose a trigonometric ratio, which involves both AE and
DE. Let us choose the tangent of the angle of elevation, ~
N tan 45° = A oo . .

ow, an = DE- 5 |
i. BE |
nee ~ 28.5 . os
Therefore, a AE = 28.5
4 "
So the height of the chimney (AB) = (28.5 + 1.5) m= 30 m.
Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m
tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation
of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the
distance of the building from the point P. (You may take ./3 = 1.732)
Solution : In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P
the given point. Note that there are two right triangles PAB and PAD. We are required
to find the length of the flagstaff, i.e, DB and the distance of the building from the
point P, i.e., PA.
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138 MATHEMATICS
Since, we know the height of the building AB, we Dg
will first consider the right A PAB. -
AB =:

We have tan 30°= -- eo
. 1 10 Zk RS
a Baap pL poe ae

A
Therefore, AP = 10V3 Fig. 9.7 :
i.e., the distance of the building from P is 10V3 m= 17.32 m. .
Next, let us suppose DB = x m. Then AD = (10 + x) m. . -

_ _ AD _10tx
Now, in right A PAD, tan 45° = AP | 10/3 -
there _ lO+x oe
erefore, =e.
10V3
ie., ~x= 10 (v3~-1) =7.32
co
So, the length of the flagstaff is 7.32 m. . A
Example 5 : The shadow of a tower standing
on a level ground is found to be 40 m longer
when the Sun's altitude is 30° than when it is
60°. Find the height of the tower.
Solution : In Fig. 9.8, AB is the tower and
BC is the length of the shadow when the p ~\ /\
Sun's altitude is 60°, i.e., the angle of | <--- 40m --+C B
elevation of the top of the tower from the tip Fis. 9.8
of the shadow is 60° and DB is the length of 1B 7
the shadow, when the angle of elevation
is 30°.
Now, let AB be A m and BC be x m. According to the question, DB is 40 m longer
than BC.
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SOME APPLICATIONS OF TRIGONOMETRY 139
So, DB = (40+x)m
Now, we have two right triangles ABC and ABD.
In AABC tan 60° A

n , an = Be

h
or, V3 = (1)
x

In AABD 30° AB

n , tan = Bp '

. 1 h 5
Le., V3 > x+40 ©
From (1), we have h= xJ3 . .
Putting this value in (2), we get (xV/3)V3 =x+40,ie., 3x=x440
iLe., x= 20 oO, / :

So, h= 20V3_ - OO [From (1)]
Therefore, the height of the tower is 20/3 m.
Example 6 : The angles of depression of the top and the bottom of an 8 m tall building
from the top of a multi-storeyed building are 30°

and 45°, respectively. Find the height of the multi-

storeyed building and the distance between the Q------------------y--g--,P
two buildings. 30°§5
Solution : In Fig. 9.9, PC denotes the multi-

storyed building and AB denotes the 8 m tall
building. We are interested to determine the height
of the multi-storeyed building, i.c., PC and the B D
distance between the two buildings, i.e., AC.
Look at the figure carefully. Observe that PB is

a transversal to the parallel lines PQ and BD. A c
Therefore, ZQPB and ZPBD are alternate

angles, and SO are equal. Fig. 9.9

So Z PBD = 30°. Similarly, Z PAC = 45°,
In right A PBD, we have

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140 MATHEMATICS

PD 1

-- = c= BD = PD

BD = (an 30 B or V3
In right A PAC, we have

PC 45° 1

AC > tan =
L.e., PC = AC
Also, PC = PD + DC, therefore, PD + DC = AC. :
Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD= PDV3 (Why?) -
This gives PD = --- = ---->_ = 4 (v3 + 1)m.

V3-1 (V3 +1)(V3-1) =, :
So, the height of the multi-storeyed building is {4(v3 + 1) + stm = 4(3 + V3)m
and the distance between the two buildings is also 4(3 + V3) m.
Example 7 : From a point on a bridge across a river, the angles of depression of
the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge
is at a height of 3 m from the banks, find the width of the river.
oe N
Solution ; In Fig 9.10, A and B ° P
represent points on the bank on . St TAEE PANY
opposite sides of the river, so that RE am NG -
AB is the width of the river. Pisa (QuE2F SREY J E SES
point on the bridge at a height of 3 A DB
m, i.e., DP = 3 m.* We are ;
interested to determine the width Fig. 9.10
of the river, which is the length of
the side AB of the D APB.
Now, _AB = AD + DB
In right AAPD, Z A= 30°.
S tan 30° "
0, an = AD
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SOME APPLICATIONS OF TRIGONOMETRY 141
1 3
Le., Fe = zn or AD= 3V3 m
V3 ~ AD N3
Also, in right A PBD, Z B = 45°. So, BD = PD =3 m.
Now, AB = BD+AD=3 + 3V¥3 =3(1+ V3) m.
Therefore, the width of the river is 3 (3 + 1)m,
EXERCISE 9.1 |

1. Acircus artist is climbing a 20 m long rope, which is A
tightly stretched and tied from the top of a vertical -
pole to the ground. Find the height of the pole, if 20m
the angle made by the rope with the ground level is
30° (see Fig. 9.11). ,

2. A tree breaks due to storm and the broken part BO (™. Cc
bends so that the top of the tree touches the ground -_.
making an angle 30° with it. The distance between . Fig. 9.11
the foot of the tree to the point where the top
touches the ground is 8 m. Find the height of
the tree. / oy

3. A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and
is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have
a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case?

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m
away from the foot of the tower, is 30°. Find the height of the tower.

5. A kite is flying ata height of 60 m above the ground. The string attached to the kite is
temporarily tied to a point on the ground. The inclination of the string with the ground
is 60°. Find the length of the string, assuming that there is no slack in the string.

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance he walked towards the building.

7. From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.
Find the height of the tower.

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