Question

In a triangle $ABC$ If the angles are in AP and $b:c=\sqrt{3}:\sqrt{2}$ then $\angle A=$

• A. ${60}^{\circ }$
• B. ${75}^{\circ }$
• C. ${15}^0$
• D. ${90}^0$

Answer 1

The correct option is B ${75}^{\circ }$

Given In $△ABC,$angles are in $A.P$

$⟹2\angle B=\angle A+\angle C$

$⟹\angle B={60}^{\circ }$

$b:c=\sqrt{3}:\sqrt{2}⟹\sin B:\sin C=\sqrt{3}:\sqrt{2}⟹\frac{\sin {60}^{\circ }}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}⟹\sin C=\frac{1}{\sqrt{2}}⟹C={45}^{\circ }$

$\angle A={120}^{\circ }-\angle C={120}^{\circ }-{45}^{\circ }={75}^{\circ }$