Question

If in a triangle $ABC$ side $a=(\sqrt{3}+1)$ cm and $\angle B={30}^0,\angle C={45}^0$, then the area of the triangle is

• A. $\frac{\sqrt{3}+1}{3}{cm}^2$
• B. $\frac{\sqrt{3}+1}{2}{cm}^2$
• C. $\frac{\sqrt{3}+1}{2\sqrt{2}}{cm}^2$
• D. $\frac{\sqrt{3}+1}{3\sqrt{2}}{cm}^2$

Answer 1

The correct option is B $\frac{\sqrt{3}+1}{2}{cm}^2$

Given, $a=(\sqrt{3}+1);\angle B={30}^0,\angle C={45}^0$

$\Rightarrow \angle A={180}^0-{75}^0={105}^0$

By sine formula, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin {45}^0}$

$\Rightarrow \frac{\sqrt{3}+1}{\sin {105}^0}=\frac{b}{\sin {30}^0}=\frac{c}{\sin {45}^0}$

$\Rightarrow b=\frac{(\sqrt{3}+1)\sin {30}^0}{\sin {105}^0}=\frac{(\sqrt{3}+1)}{2\sin ({45}^0+{60}^0)}=\sqrt{2}$

$\therefore$ Area of $\Delta =\frac{1}{2}ab\sin C$

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