Question

The sides of a triangle are in the ratio $2:\sqrt{6}:(\sqrt{3}+1)$, the angles are ${45}^o,{60}^o,{75}^o$. (Enter $1$ if true or $0$ otherwise)

Answer 1

We know from sine Rule

$\frac{a}{\sin \pi }=\frac{b}{\sin B}=\frac{C}{\sin C}=2R$

$\frac{a=2}{\sin 45}=\frac{\sqrt{6}}{\sin 60}=\frac{\sqrt{3}+1}{\sin 75}=2R$

$\frac{2}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}+1}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=2R$

$2\sqrt{2}=2\sqrt{2}=2\sqrt{2}=2R$

Hence proved