Trigonometric Equations Question 112
Question: In a triangle $ ABC $ , if $ B=3C $ , then the values of $ \sqrt{( \frac{b+c}{4c} )} $ and $ ( \frac{b-c}{2c} ) $ are
Options:
A) $ \sin C,\sin \frac{A}{2} $
B) $ \cos C,\sin \frac{A}{2} $
C) $ \sin C,\cos \frac{A}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \sqrt{\frac{b+c}{4c}}=\sqrt{\frac{\sin 3C+\sin C}{4\sin C}} $
Þ $ \sqrt{\frac{2\sin 2C\cos C}{4\sin C}}=\cos C $ $ \frac{b-c}{2c}=\frac{\sin 3C-\sin C}{2\sin C}=\frac{2\cos 2C\sin C}{2\sin C}=\cos 2C=\sin \frac{A}{2} $ .
BETA
