JEE Main · 2012 · 29 Apr 2012 · Offline · Properties of Triangle

Square and add the two given relations: $(3\sin P+4\cos Q{)}^2+(4\sin Q+3\cos P{)}^2=36+1=37$. Expanding and grouping the squares, $9+16+24(\sin P\cos Q+\cos P\sin Q)=37$, so $24\sin (P+Q)=12$ and hence $\sin (P+Q)=\frac{1}{2}$. The first equation can only hold if $\sin P$ is large, which forces $P$ to be obtuse, so $P+Q=\frac{5\pi }{6}$. The remaining angle is $R=\pi -(P+Q)=\frac{\pi }{6}$.