Question

The perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles.If the side $a$ is $1$, then $\angle A$ is

• A. ${30}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${120}^0$

Answer 1

The correct option is A ${30}^0$

Let $\sin A,\sin B,\sin C$ be the three angles.
Then Given:Perimeter of $△ABC=6\times$A.M of sines of its angles.
$\Rightarrow 2s=6\times \left(\frac{\sin A+\sin B+\sin C}{3}\right)$
$\therefore s=\sin A+\sin B+\sin C$ (on simplification)
Using sine rule we have
$s=ak+bk+ck$
$\Rightarrow s=(a+b+c)k$
$\Rightarrow s=\left(2s\right)k$
$\therefore 1=2k$ or $k=\frac{1}{2}$
$\Rightarrow \sin A=ak=1\times \frac{1}{2}$
$\therefore \angle A={30}^0$