Question

In a $\triangle A B C$, if $\angle A=\angle B=\frac{1}{2}\left(\sin ^{-1}\left(\frac{\sqrt{6}+1}{2 \sqrt{3}}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$ and $c=6 \cdot 3^{\frac{1}{4}}$, then find the area of $\triangle ABC$.

Answer 1

$ x^2=12-(7+2 \sqrt{5})=5-2 \sqrt{6}$
$x ^2=(\sqrt{3}-\sqrt{2})^2 \Rightarrow x =(\sqrt{3}-\sqrt{2})$
Hence $\theta=\tan ^{-1}\left(\frac{1+\sqrt{2} \sqrt{3}}{\sqrt{3}-\sqrt{2}}\right)+\tan ^{-1} \frac{1}{\sqrt{2}}$

$\theta =\frac{\pi}{2}-\tan ^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{2} \sqrt{3}}\right)+\tan ^{-1} \frac{1}{\sqrt{2}}=\frac{\pi}{2}-\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{2}\right)+\cot ^{-1} \sqrt{2} $
$ =\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$
Hence $\angle A =\angle B =\frac{1}{2}\left(\frac{2 \pi}{3}\right)=\frac{\pi}{3}$
$\therefore \angle C =\frac{\pi}{3}$
So, $\triangle ABC$ is equilateral.
Hence area $(\triangle ABC )=\frac{\sqrt{3}}{4} c ^2=\frac{\sqrt{3}}{4}\left(6 \cdot 3^{\frac{1}{4}}\right)^2=27$