Question

The angles A, B and C of a triangle ABC are in AP, if b : c = : $\sqrt{3}:\sqrt{2}$, then the angle A is

• A. 30°
• B. 15°
• C. 75°
• D. 45°

Answer 1

The correct option is C

75°

Since, A,B,C are in A.P.

$\therefore 2B=A+C\Rightarrow B={60}^0(\because A+B+C={180}^0)\text{Using sine rule,}\frac{sinB}{b}=\frac{sinC}{c}\therefore \frac{sin{60}^0}{\sqrt{3}}=\frac{sinC}{\sqrt{2}}\Rightarrow \frac{\sqrt{3}}{2\sqrt{3}}=\frac{sinC}{\sqrt{2}}\Rightarrow sinC=\frac{1}{\sqrt{2}}\Rightarrow C={45}^0\therefore A={180}^0-({60}^0+{45}^0)={75}^0$