Trigonometric Equations Question 218
Question: In triangle $ ABC $ if $ a,b,c $ are in A. P., then the value of $ \frac{\sin \frac{A}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}}= $
[AMU 1995]
Options:
A) 1
B) 1/2
C) 2
D) -1
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \frac{\sin \frac{A}{2}\sin \frac{C}{2}}{\sin \frac{B}{2}}=\sqrt{\frac{ac(s-b)(s-c)(s-b)(s-a)}{(s-a)(s-c)bc\times ab}}=\frac{s-b}{b} $ But a, b and c are in A. P. Þ $ 2b=a+c $ Hence $ \frac{s-b}{b}=\frac{\frac{3b}{2}-b}{b}=\frac{1}{2} $ .
BETA
