Question

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

Answer 1

We have, $\tan \frac{A-B}{2}=\sqrt{\frac{1-\cos (A-B)}{1+\cos (A-B)}}=\sqrt{\frac{1-\frac{31}{32}}{1+\frac{31}{32}}}=\frac{1}{\sqrt{63}}$

$\Rightarrow \frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}[\because \tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}]$

$\Rightarrow \frac{1}{9}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$

Now, $\cos C=\frac{1-{\tan }^2\frac{C}{2}}{1+{\tan }^2\frac{C}{2}}\Rightarrow \cos C=\frac{1-\frac{7}{9}}{1+\frac{7}{9}}=\frac{1}{8}$

Since $c^2=a^2+b^2-2ab\cos C$

$\Rightarrow c^2=25+16-40\times \frac{1}{8}=36$

$\therefore c=6$