Trigonometric Equations Question 169
Question: ABC is a triangle such that $ \sin (2A+B)= $ $ \sin (C-A)= $ $ -\sin (B+2C)=\frac{1}{2} $ . If A, B and C are in A.P., then A, B and C are
Options:
A) $ 30^{o},60^{o},90^{o} $
B) $ 45^{o},60^{o},75^{o} $
C) $ 45^{o},45^{o},90^{o} $
D) $ 60^{o},60^{o},60^{o} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Since A,B,C are in A.P., therefore $ B=60^{o} $ [ $ \because A+B+C=180 $ and $ A+C=2B $ ] Now, $ \sin (2A+B)=\frac{1}{2} $ (given) Þ $ 2A+B=30^{o} $ or $ 150^{o} $ But as $ B=60^{o}, $ $ 2A+B\ne 30^{o} $ . Hence $ 2A+B=150^{o}\Rightarrow A=45^{o} $ Hence $ A=45^{o} $ , $ B=60^{o},C=75^{o} $ .
BETA
