Discussion Forum

 With usual notations, if the angles A,B,C of a  $\triangle$ABC  are in AP  and b:c= $\sqrt{3}:\sqrt{2}$

Answer:

Explanation:

In  $\triangle$ABC,

$\angle A,\angle B, \angle C$ are in A.P

 $\therefore$           2$\angle B$=  $\angle A+ \angle C$.............(i)

 and   $\angle A+\angle B+ \angle C =180^{0}$..........(ii)

  From eqs.(i) and (ii) , B= 60°

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

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

 $\Rightarrow\frac{\sin 60^{0}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}$     $[\because$  given   $b:c=\sqrt{3}:\sqrt{2}]$

$\Rightarrow \sin C=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times2}=\frac{1}{\sqrt{2}}$

 $\therefore$        sin C= sin  45°

 $\Rightarrow$       C=45°

 $\angle A=2\angle B-\angle C=120^{0}-45^{0}=75^{0}$