Question

Angles $A,B$ and $C$ of a triangle $ABC$ are in $A.P$. If $\frac{b}{c}=\sqrt{\frac{3}{2}}$, then $\angle A$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{5\pi }{12}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{5\pi }{12}$

As the angles are in A.P., so, let

$\angle A=x-d,\angle B=x,\angle C=x+d$

Since the sum of the angles of a triangle is ${180}^{\circ }$, so,

$x-d+x+x+d={180}^{\circ }$

$x={60}^{\circ }$

$\angle B={60}^{\circ }$

By sine law,

$\frac{b}{\sin B}=\frac{c}{\sin C}$

$\frac{\sin B}{\sin C}=\frac{b}{c}$

$\frac{\sin {60}^{\circ }}{\sin C}=\sqrt{\frac{3}{2}}$

$\frac{\sqrt{3}}{2\sin C}=\frac{\sqrt{3}}{\sqrt{2}}$

$\sin C=\frac{1}{\sqrt{2}}$

$\sin C=\sin {45}^{\circ }$

$C={45}^{\circ }$

$\angle A={180}^{\circ }-\left(\angle B+\angle C\right)$

$\angle A={180}^{\circ }-\left({60}^{\circ }+{45}^{\circ }\right)$

$\angle A={75}^{\circ }$