NCERT Class 10 Mathematics

INTRODUCTION TO TRIGONOMETRY 179

Then

AC

AB

=

PR

PQ

Therefore,

AC

PR

=

AB

,say

PQ

k (1)

Now, using Pythagoras theorem,

BC = AB^22 ✁AC

and QR = PQ – PR^22

So,

BC

QR =

2 2 22 22 2 2

2 2 22 22

AB AC PQ PR PQ PR

PQ PR PQ PR PQ PR

kk k

k

✂ ✂ ✂

✄ ✄ ✄

✂ ✂ ✂

(2)

From (1) and (2), we have

AC

PR

=

AB BC

PQ QR

Then, by using Theorem 6.4, ☎ ACB ~ ☎ PRQ and therefore, ✆ B = ✆ Q.

Example 3 : Consider ☎ ACB, right-angled at C, in

which AB = 29 units, BC = 21 units and ✆ ABC = ✝

(see Fig. 8.10). Determine the values of

(i) cos^2 ✝ + sin^2 ✝,

(ii) cos^2 ✝ – sin^2 ✝✞

Solution : In ☎ ACB, we have

AC = AB^22 ✁BC = (29)^22 ✁(21)

= (29✟21) (29✠21)✡ (8) (50)✡ 400 ✡20 units

So, sin ✝ =

AC (^20) , BC 21
cos =
AB 29 AB 29

☛ ☞ ☛ ✌

Now, (i) cos^2 ✝ + sin^2 ✝ =

(^2222)
2

20 21 20 21 400 441

1,

29 29 29 841

✎ ✏ ✎ ✏ ✍ ✍

✒ ✓ ✍✒ ✓ ✑ ✑ ✑

✔ ✕ ✔ ✕

and (ii) cos^2 ✝ – sin^2 ✝ =

22

2

21 20 (21 20) (21 20) 41

29 29 29 841

✎ ✏ ✎ ✏ ✍ ✖

✒ ✓ ✖✒ ✓ ✑ ✑

✔ ✕ ✔ ✕

.

Fig. 8.10