Trigonometric Equations Question 97
Question: In a triangle $ ABC $ , $ \tan \frac{A}{2}=\frac{5}{6} $ and $ \tan \frac{C}{2}=\frac{2}{5}, $ then
[EAMCET 1994]
Options:
A) $ a,\ b,\ c $ are in A.P.
B) $ \cos A,\ \cos B,\ \cos C $ are in A.P.
C) $ \sin A,\ \sin B,\ \sin C $ are in A.P.
D) (a) and (c) both
Show Answer
Answer:
Correct Answer: D
Solution:
- Here $ \tan \frac{A}{2}\tan \frac{C}{2}=\frac{s-b}{s} $ $ \frac{5}{6}.\frac{2}{5}=\frac{s-b}{s}\Rightarrow 3s-3b=s\Rightarrow 2s=3b $
$ \Rightarrow $ $ a+b+c=3b $ or $ a+c=2b $ .
$ \therefore $ a, b, c are in A.P., also sinA, sinB, sinC are in A.P.
BETA
