Question

If in a triangle the angles are in A.P. and $b:c=\sqrt{3}:\sqrt{2}$

then $△A$ is equal to

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

Answer 1

The correct option is D ${75}^o$

A, B, C are in A.P

and $b:c=\sqrt{3}:\sqrt{2}$

To find: $\angle A$

Let $\angle A=a-d,\angle B=a$,

$\angle C=a+d$

$a-d+a+a+d={180}^o$(sum of angles of triangle)

$3a={180}^o$

$a={60}^o$

$\therefore \angle B={60}^o$

using sine law,

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

$\frac{\sin {60}^o}{\sin c}=\frac{b}{c}$

$\frac{\sqrt{3}/2}{\sin c}=\frac{\sqrt{3}}{\sqrt{2}}$

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

$\sin c=\sin {45}^o$

$\angle c={45}^o$

$a+d={45}^o$

$d={45}^o-{60}^o=-{15}^o$

$\angle A=a-d={60}^o-(-{15}^o)$

$\therefore \angle A={75}^o$.