Question

In $\Delta PQR$, let $\angle P>\angle Q$. If the radian measures of $\angle P$ and $\angle Q$ satisfy the equation $4{\sin }^{3}x-3\sin x+a=0,0<a<1$, then the radian measure of $\angle R$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (C)

Given, $4{\sin }^{3}x-3\sin x+a=0$
$\Rightarrow a=3\sin x-4{\sin }^{3}x$
$\Rightarrow a=\sin 3x$
$\because \sin 3P=a=\sin 3Q$
[since $P$ and $Q$ satisfy the equation]
$\Rightarrow \sin 3P=\sin 3Q$
$\Rightarrow \sin 3P=\sin (\pi -3Q)$
$\Rightarrow 3P=\pi -3Q$
$\Rightarrow 3(P+Q)=\pi$
$\Rightarrow dP+Q=\frac{\pi }{3}$
$\because \angle P+\angle Q+\angle R=\pi$
$\therefore \angle R=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$