Question

In a triangle $ABC,$ if $A,B,C$ are in A.P and $b:c=\sqrt{3}:\sqrt{2}$, then $A$

• A. ${45}^0$
• B. ${60}^0$
• C. ${75}^0$
• D. ${30}^0$

Answer 1

The correct option is C ${75}^0$

$A,B,C$ are in A.P.
$\Rightarrow A+C=2B$
$\Rightarrow \pi -B=2B$ ($\because A+B+C=\pi$)
$\Rightarrow 3B=\pi$ or $B=\frac{\pi }{3}={60}^0$
Given $b:c=\sqrt{3}:\sqrt{2}$
$\Rightarrow \frac{\sin B}{\sin C}=\sqrt{3}:\sqrt{2}$
$\Rightarrow \frac{\sin {60}^0}{\sin C}=\sqrt{3}:\sqrt{2}$
$\Rightarrow \frac{\frac{\sqrt{3}}{2}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}$
$\Rightarrow 2\sin C=\sqrt{2}$
$\Rightarrow \sin C=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{\sqrt{2}}$ (by rationalizing the numerator)
On simplifying we get
$\Rightarrow \sin C=\frac{1}{\sqrt{2}}$
$\therefore C={45}^0$
$A={180}^0-(B+C)={180}^0-({60}^0+{45}^0)={75}^0$ (on simplification)
Hence,$\angle A={75}^0$