Question

In a $△ABC,$ if $a,b,c$ are the sides opposite to angles $A,B,C$ respectively and $a=6,b=3,$ $\cos (A-B)=\frac{4}{5},$ then

• A. $C=\frac{\pi }{4}$
• B. $C=\frac{\pi }{2}$
• C. $C=\frac{\pi }{6}$
• D. $C=\frac{\pi }{3}$

Answer 1

The correct option is B $C=\frac{\pi }{2}$

$\cos (A-B)=\frac{4}{5}$
$\Rightarrow \frac{1-{\tan }^2\left(\frac{A-B}{2}\right)}{1+{\tan }^2\frac{A-B}{2}}=\frac{4}{5}[\because \cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }]$
$\Rightarrow {\tan }^2\frac{A-B}{2}=\frac{1}{9}$
$\Rightarrow \tan \frac{A-B}{2}=\pm \frac{1}{3}$
By Napier’s analogy
$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$
$\Rightarrow \pm \frac{1}{3}=\frac{3}{9}\cot \frac{C}{2}$
$\Rightarrow \cot \frac{C}{2}=1\{\because \frac{C}{2}\in (0,\frac{\pi }{2})\}$
$\therefore C=\frac{\pi }{2}$