Question

In a $△ABC$, if $a=5,b=4$ and $\cos (A-B)=\frac{31}{32}$, then side $c$ is

• A. $6$
• B. $7$
• C. $9$
• D. None of the above

Answer 1

Answer: (A)

$6$

Given, $a=5,b=4,\cos (A-B)=\frac{31}{32}$
Therefore, $\tan \left(\frac{A-B}{2}\right)=\sqrt{\frac{1-\cos (A-B)}{1+\cos (A+B)}}=\sqrt{\frac{1}{63}}$
$\Rightarrow \frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \frac{5-4}{5+4}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}$
$\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$
Now, $\cos C=\frac{1-{\tan }^2\frac{C}{2}}{1+{\tan }^2\frac{C}{2}}$
$\Rightarrow \cos C=\frac{1-\frac{7}{9}}{1+\frac{7}{9}}=\frac{1}{8}$
Now, $c^2=a^2+b^2-2ab\cos C$
$\Rightarrow c^2=25+16-40\times \frac{1}{8}=36$
$\Rightarrow c=6$