Question

In a $△ABC$, if $\cos A\cos B\cos C=\frac{\sqrt{3}-1}{8}$ and $\sin A\sin B\sin C=\frac{3+\sqrt{3}}{8}$, then the angles of the triangle are

• A. ${45}^{\circ },{30}^{\circ },{105}^{\circ }$
• B. ${45}^{\circ },{60}^{\circ },{75}^{\circ }$
• C. ${45}^{\circ },{45}^{\circ },{90}^{\circ }$
• D. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ },{60}^{\circ },{75}^{\circ }$

$\because A,B,C$ are angles of triangle,
$\therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C=\frac{\sin A\sin B\sin C}{\cos A\cos B\cos C}=\frac{3+\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=2\sqrt{3}+3$
By observation,
$1\cdot \sqrt{3}\cdot (2+\sqrt{3})=1+\sqrt{3}+2+\sqrt{3}$
So,
$\tan A,\tan B,\tan C$ are $1,\sqrt{3},2+\sqrt{3}$
Therefore, the angles are
${45}^{\circ },{60}^{\circ },{75}^{\circ }$