Question

The angles of a $\Delta ABC$ are in A.P. (order being $A,B,C$) and it is being given that $b:c=\sqrt{3}:\sqrt{2}$, then find $\angle A$.

Answer 1

Let A$=x-d$, B$=x$, C$=x+d$ are angles of triangle

$x-x+x+x+d=180$

$3x=180\Rightarrow x={60}^o\left[\right(B\left)\right]$

We know $\frac{b}{\sin B}=\frac{c}{\sin C}\to \frac{b}{c}=\frac{\sin B}{\sin C}$

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

$\Rightarrow (\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin 60}{\sin C})$ $\Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{\frac{2}{\sin C}}$

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

$\Rightarrow C={45}^o$

$\angle A=180-B-C$

$\Rightarrow 180-60-45={75}^o$

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