Trigonometric Equations Question 301
Question: In a $ \Delta ABC $ if the sides are $ a=3,,b=5 $ and $ c=4 $ , then $ \sin \frac{B}{2}+\cos \frac{B}{2} $ is equal to
[Karnataka CET 2005]
Options:
A) $ \sqrt{2} $
B) $ \frac{\sqrt{3}+1}{2} $
C) $ \frac{\sqrt{3}-1}{2} $
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
- $ a=3,b=5,c=4,s=\frac{a+b+c}{2}=\frac{12}{2}=6 $ $ \sin \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}}=\sqrt{\frac{2.3}{12}}=\sqrt{\frac{1}{2}} $ $ \cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}}=\sqrt{\frac{6.1}{12}}=\sqrt{\frac{1}{2}} $ \ $ \sin \frac{B}{2}+\cos \frac{B}{2}=\frac{2}{\sqrt{2}}=\sqrt{2} $ .
BETA
