Question

If angles $A,B$ and $C$ of a triangle $ABC$ are in A.P,if $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$,then find angle $A$

Answer 1

Given $A,B,C$ in A.P

Let $A=B-d,B=B,C=B+d$

$A+B+C=\pi \Rightarrow B-d+B+B+d={180}^0\Rightarrow 3B=180$

$\Rightarrow B={60}^0$

Now using sine rule, $\frac{\sin B}{b}=\frac{\sin C}{c}$

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

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

$\Rightarrow \sin C=\frac{1}{\sqrt{2}}=\sin {45}^0$

$\Rightarrow C={45}^0$

$\Rightarrow A+B+C={180}^0\Rightarrow A+45+60=180\Rightarrow A={75}^0$

$\therefore A={75}^0$