Question

In a $△ABC$ with usual notation, if $\angle A=\angle B=\frac{1}{2}({\sin }^{-1}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right))$ and $c=6\cdot 3^{1/4},$ then the area of $△ABC$ in square units, is

Answer 1

Let $\sin \theta =\frac{\sqrt{6}+1}{2\sqrt{3}}$


$2A=2B={\sin }^{-1}\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\therefore 2A=2B={\tan }^{-1}\left(\frac{1+\sqrt{6}}{\sqrt{3}-\sqrt{2}}\right)+{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$={\cot }^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{3}\cdot \sqrt{2}}\right)+{\tan }^{-1}\frac{1}{\sqrt{2}}$
$=\frac{\pi }{2}-[{\tan }^{-1}\sqrt{3}-{\tan }^{-1}\sqrt{2}]+{\tan }^{-1}\frac{1}{\sqrt{2}}$
$=\frac{\pi }{2}-{\tan }^{-1}\sqrt{3}+{\tan }^{-1}\sqrt{2}+{\cot }^{-1}\sqrt{2}$
$=\pi -{\tan }^{-1}\sqrt{3}=\frac{2\pi }{3}$
$\Rightarrow 2A=2B=\frac{2\pi }{3}$
$\Rightarrow A=B=\frac{\pi }{3}$
$\Rightarrow \angle C=\frac{\pi }{3}$
$\therefore △ABC$ is equilateral.
Hence, area $=\frac{\sqrt{3}}{4}\times c^2=\frac{\sqrt{3}}{4}\times 36\times \sqrt{3}=27$ sq. units