Question

The angles of the ΔABC are in A.P. and b:c=${\displaystyle \sqrt{3}:\sqrt{2}}$ then find${\displaystyle \angle A,\angle B,\angle C}$

 

Answer 1

${\displaystyle \angle A,\angle B,\angle C}$ are in A.P and ${\displaystyle b:c=\sqrt{3}:\sqrt{2}}$

${\displaystyle \therefore 2B=A+C}$

${\displaystyle 2B={180}^{\circ }-B}$     ...............[${\displaystyle \because A+B+C={180}^{\circ }}$]

${\displaystyle 3B={180}^{\circ }}$

${\displaystyle \angle B={60}^{\circ }}$

In ΔABC by sine rule, we have

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

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

${\displaystyle \frac{{\sin 60}^{\circ }}{\sin c}=\frac{\sqrt{3}}{\sqrt{2}}}$

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

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

${\displaystyle \angle C={45}^{\circ }}$

${\displaystyle \angle A={180}^{\circ }-{60}^{\circ }-{45}^{\circ }={75}^{\circ }}$

Thus, the angles of ΔABC are ${\displaystyle \angle A={75}^{\circ },\angle B={60}^{\circ },\angle C={45}^{\circ }}$