Maths

INTRODUCTION TO TRIGONOMETRY 179
Then
Therefore,
Now, using Pythagoras theorem,
AC
AB = PR
PQ
AC
PR = AB ,say
PQ k (1)
2 2
BC = AB AC
2 2
and QR = PQ – PR
✁
So,
2 2 2 2 2 2 2 2
BC AB AC k PQ k PR k PQ PR
✂ ✂ ✂
QR = 2 2 2 2 2 2
PQ PR PQ PR PQ PR
✄ ✄
✄
✂ ✂
✂
From (1) and (2), we have
AC
PR = AB BC
PQ QR
Then, by using Theorem ☎ 6.4, ACB ☎ ~ PRQ and ✆ therefore, B ✆ = Q.
k
(2)
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
2 2
AC = AB BC
(29) (21)
✁ =
2 2
✁
Fig. 8.10
✟ ✠ ✡ ✡ ✡
= (29 21)(29 21) (8)(50) 400 20units
AC 20 BC 21
So, sin = , ☛ cos ☞ ✝ ☛ =
✌
AB 29 AB 29
Now, (i) cos 2 ✝ + sin 2 ✝ =
and (ii) cos 2 ✝ – sin 2 ✝ =
2 2 2 2
20 21 20 21 400 441
✍
✍
✎ ✏ ✎ ✏
✍ ✑ ✑ ✑
✒ ✓ ✒ ✓
2
✕ ✔ ✕
✔ 29 29 841
2 2
29
21 20 (21 20)(21 20) 41
.
29
✍ ✖
✎ ✏ ✎ ✏
✖ ✑ ✑
✒ ✓ ✒ ✓
2
✕ ✔ ✕
✔ 29 29 841
1,