Question

In a $△ABC$, 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 $△ABC$.

Answer 1

Answer: 27

$x^2=12-(7+2\sqrt{5})=5-2\sqrt{6}$
$x^2=(\sqrt{3}-\sqrt{2}{)}^2\Rightarrow x=(\sqrt{3}-\sqrt{2})$
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, $△ABC$ is equilateral.
Hence area $(△ABC)=\frac{\sqrt{3}}{4}{c}^2=\frac{\sqrt{3}}{4}{\left(6\cdot 3^{\frac{1}{4}}\right)}^2=27$