Trigonometric Equations Question 247
Question: In a $ \Delta ABC $ , if $ A=30^{o} $ $ b=2,c=\sqrt{3}+1 $ , then $ \frac{C-B}{2}= $
Options:
A) $ 15^{o} $
B) $ 30^{o} $
C) $ 45^{o} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- We have $ \tan \frac{C-B}{2}=\frac{c-b}{c+b}\cot \frac{A}{2} $ \ $ \tan ( \frac{C-B}{2} )=\frac{\sqrt{3}+1-2}{\sqrt{3}+1+2}\cot 15^{o}=\frac{1}{\sqrt{3}} $
Þ $ \frac{C-B}{2}=30^{o} $ .
BETA
