Question

If the angle of a triangle are ${30}^{\circ }$ and ${45}^{\circ },$ and the included side is $\sqrt{3}+1$.Prove that the area of the triangle is $\frac{\sqrt{3}+1}{2}$

Answer 1

By angle sum property,${30}^{\circ }+{45}^{\circ }+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

We have $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\Rightarrow \frac{\sin {105}^{\circ }}{\sqrt{3}+1}=\frac{\sin {30}^{\circ }}{b}=\frac{\sin {45}^{\circ }}{c}$

$\Rightarrow b=\frac{\sqrt{3}+1}{2\sin {105}^{\circ }}$ and $c=\frac{\sqrt{3}+1}{\sqrt{2}\sin {105}^{\circ }}$

So, area of $△ABC=\frac{1}{2}bc\sin A$

$=\frac{1}{2}\times \frac{{(\sqrt{3}+1)}^2}{2\sqrt{2}{\sin }^2\left({105}^{\circ }\right)}\times \sin \left({150}^{\circ }\right)$

$=\frac{1}{2}\times \frac{{(\sqrt{3}+1)}^2}{2\sqrt{2}\sin ({60}^{\circ }+{45}^{\circ })}$

$\sin {105}^{\circ }=\sin ({60}^{\circ }+{45}^{\circ })=\sin {60}^{\circ }\cdot \cos {45}^{\circ }+\cos {60}^{\circ }\cdot \sin {45}^{\circ }=\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}+\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$=\frac{1}{2}\frac{{(\sqrt{3}+1)}^2}{2\sqrt{2}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}$

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