Trigonometric Equations Question 233
Question: In a $ \Delta ABC $ , $ a=5,b=4 $ and $ \cos (A-B)=\frac{31}{32} $ , then side c is equal to
Options:
A) 6
B) 7
C) 9
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ \tan ( \frac{A-B}{2} )=\sqrt{\frac{1-\cos (A-B)}{1+\cos (A-B)}} $ $ =\sqrt{\frac{1-(31/32)}{1+(31/32)}}=\frac{1}{\sqrt{63}} $
Þ $ \frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{\sqrt{63}} $
Þ $ \frac{1}{9}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3} $ Now $ \cos C=\frac{1-{{\tan }^{2}}(C/2)}{1+{{\tan }^{2}}(C/2)} $ Þ $ \cos C=\frac{1-(7/9)}{1+(7/9)}=\frac{1}{8} $ \ $ c^{2}=a^{2}+b^{2}-2ab\cos C $
Þ $ c^{2}=25+16-40\times \frac{1}{8}=36\Rightarrow c=6 $ .
BETA
