Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Squaring and adding: $9+16+24\sin (P+Q)=37\Rightarrow \sin (P+Q)=\frac{1}{2}\Rightarrow P+Q=\frac{\pi }{6}$ or $\frac{5\pi }{6}$. If $P+Q=\frac{\pi }{6}$, then $P<\frac{\pi }{6},3\sin P<\frac{3}{2}$ and $4\cos Q\le 4$, so $3\sin P+4\cos Q<\frac{3}{2}+4<6$, which is not possible. So $P+Q=\frac{5\pi }{6}$ and $R=\frac{\pi }{6}$.