Question

In a $△ABC,a=2b$ and $|A-B|=\frac{\pi }{3}$.Then the $\angle C$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. none of the above

Answer 1

The correct option is C $\frac{\pi }{3}$

Here,$A>B$
$\therefore \tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot \frac{C}{2}$
From the question, $\tan {30}^0=\frac{2b-b}{2b+b}\cot \frac{C}{2}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{3}\cot \frac{C}{2}$
$\Rightarrow \cot \frac{C}{2}=\frac{3}{\sqrt{3}}$
$\Rightarrow \cot \frac{C}{2}=\sqrt{3}$
$\therefore \frac{C}{2}=\frac{\pi }{6}$
or $C=\frac{\pi }{3}$