Question

In $△ABC,$ and $|A-B|=\frac{\pi }{3}$.Determine the angle $C$.

Answer 1

Here $a=2b$

$\angle A>\angle B$ as $a>b$ and we know greater angle is opposite to greater side.

$\Rightarrow |A-B|=\frac{\pi }{3}$

or $A-B=\frac{\pi }{3}$

using Napier's analogy, we have

$\tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot \left(\frac{C}{2}\right)$

$\Rightarrow \tan \frac{\pi }{6}=\frac{2b-b}{2b+b}\cot \left(\frac{C}{2}\right)$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{3}\cot \frac{C}{2}$

$\Rightarrow \cot \frac{C}{2}=\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}$

$\therefore C=\frac{\pi }{3}$