Question

Let the area of triangle ABC be $(\sqrt{3}-1)/2,b=2,$ and $c=(\sqrt{3}-1),$ and $\angle A$ be acute. Then measure of the angle $A$ is

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

Answer: (B)

${30}^{\circ }$

Given : Area of triangle ABC $=\frac{\sqrt{3}-1}{2}$

Area of triangle will be $=\frac{bc\sin A}{2}$
$⟹\frac{\sqrt{3}-1}{2}=\frac{2(\sqrt{3}-1).\sin A}{2}$
$⟹\frac{\sqrt{3}-1}{2}=(\sqrt{3}-1)\sin A$
$⟹\sin A=\frac{1}{2}$
$⟹A={30}^0$.