Question

In $△ABC,a=5,b=4$ and $\cos (A-B)=\frac{31}{32}$ find $C$

Answer 1

Using $\cos \left(\frac{A-B}{2}\right)=\frac{1-{\tan }^2\frac{A-B}{2}}{1+{\tan }^2\frac{A-B}{2}}=\frac{31}{32}\Rightarrow \tan \frac{A-B}{2}=\frac{1}{\sqrt{63}}$

Using $\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{63}}{9}$

Using $\cos C=\frac{1-{\tan }^2\frac{C}{2}}{1+{\tan }^2\frac{C}{2}}\Rightarrow \cos C=\frac{1}{8}$

Using $c^2=a^2+b^2-2ab\cos C=36$

$C=6$