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