Question

In a $\Delta ABC,a=(1+\sqrt{3})cm,b=2cm$ and $\angle C={60}^{\circ }$ , then find the other two angle and the third side

• A. $\angle A={75}^{\circ },\angle B={45}^{\circ }$ , $c=\sqrt{6}$
• B. $\angle A={60}^{\circ },\angle B={60}^{\circ }$ , $c=\sqrt{3}$
• C. $\angle A={90}^{\circ },\angle B={30}^{\circ }$ , $c=\sqrt{3}$
• D. $\angle A={45}^{\circ },\angle B={75}^{\circ }$ , $c=\sqrt{6}$

Answer 1

The correct option is A $\angle A={75}^{\circ },\angle B={45}^{\circ }$ , $c=\sqrt{6}$

Given $a=1+\sqrt{3},b=2$ and $\angle C={60}^{\circ }$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow \frac{1}{2}=\frac{(1+\sqrt{3}{)}^2+4-c^2}{4(1+\sqrt{3})}$

$\Rightarrow 2+2\sqrt{3}=4+2\sqrt{3}+4-c^2$

$\Rightarrow c=\sqrt{6}$

using Sine rule
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow \frac{1+\sqrt{3}}{\sin A}=\frac{2}{\sin B}=\frac{\sqrt{6}}{\sqrt{3}/2}$

$\Rightarrow \angle A={75}^{\circ }$ and $\angle B={45}^{\circ }$
$\therefore \angle A={75}^{\circ },\angle B={45}^{\circ }$ and $c=\sqrt{6}$