Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.
If $a=5,b=4$ and $\cos (A-B)=\frac{31}{32}$, then find side $c.$

• A. $6$
• B. $8$
• C. $9$
• D. None of these

Answer 1

The correct option is A $6$

$a=5,b=4,\cos (A-B)=\frac{31}{32}$

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

$\Rightarrow \frac{-2{\tan }^2\left(\frac{A-B}{2}\right)}{2}=-\frac{1}{63}\Rightarrow {\tan }^2\left(\frac{A-B}{2}\right)=\frac{1}{63}\Rightarrow \tan \left(\frac{A-B}{2}\right)=\pm \frac{1}{3\sqrt{7}}$

But $\cos (A-B)=\frac{31}{32}\Rightarrow A-B$ is acute

$\Rightarrow \tan \left(\frac{A-B}{2}\right)=\frac{1}{3\sqrt{7}}$

$\therefore$ By Napier's analogy, $\frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{3\sqrt{7}}$

$\Rightarrow \frac{1}{9}\cot \frac{C}{2}=\frac{1}{3\sqrt{7}}\Rightarrow \cot \frac{C}{2}=\frac{3}{\sqrt{7}}\Rightarrow \cos \frac{C}{2}=\frac{3}{4}$

$\Rightarrow \cos C=2{\left(\frac{3}{4}\right)}^2-1=\frac{1}{8}\Rightarrow \frac{1}{8}=\frac{25+16-C^2}{48}\Rightarrow C^2=36\Rightarrow C=6$