Question

Angles A, B, C of a $\Delta$ABC are in AP and $b:c=\sqrt{3}:\sqrt{2},$ then the $\angle$A is given by.

• A. ${45}^o$
• B. ${60}^o$
• C. ${75}^o$
• D. ${90}^o$

Answer 1

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

Given : Angles $A,B,C$ of $△ABC$ are in AP

Let $\angle A=a-d$

$\angle B=a$

$\angle C=a+d$

We know that, sum of all angles of a triangle is ${180}^o$

$⟹\angle A+\angle B+\angle C={180}^o$ .... $\left(i\right)$

$⟹a-d+a+a+d={180}^o$

$⟹3a={180}^o⟹a={60}^o$

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

We know by sine rule

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

$⟹\frac{a}{\sin A}=\frac{b}{c}=\frac{\sin B}{\sin C}$

$⟹\frac{a}{\sin A}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin B}{\sin C}$

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

$⟹\sin C=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{2}$ .... $[\because \sin {60}^o=\frac{\sqrt{3}}{2}]$

$⟹\sin C=\frac{1}{\sqrt{2}}$

$\therefore C={45}^o$

From $\left(i\right)$

$\angle A+{45}^o+{60}^o={180}^o$

$⟹\angle A={75}^o$