In a $\Delta PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to:

$3\sin P+4\cos Q=6$ ...(1)$4\sin Q+3\cos P=1$ ...(2)Squaring and adding both equations:$(3\sin P+4\cos Q{)}^2+(4\sin Q+3\cos P{)}^2=6^2+1^2$$9{\sin }^{2}P+16{\cos }^{2}Q+24\sin P\cos Q+16{\sin }^{2}Q+9{\cos }^{2}P+24\sin Q\cos P=37$$9({\sin }^{2}P+{\cos }^{2}P)+16({\sin }^{2}Q+{\cos }^{2}Q)+24(\sin P\cos Q+\cos P\sin Q)=37$$9(1)+16(1)+24\sin (P+Q)=37$$25+24\sin (P+Q)=37$$24\sin (P+Q)=12⟹\sin (P+Q)=\frac{1}{2}$ Since $P,Q$ are angles of a triangle, $P+Q$ can be $\frac{\pi }{6}$ or $\frac{5\pi }{6}$.However, from (1), $3\sin P+4\cos Q=6$, which requires $\sin P$ and $\cos Q$ to be large. If $P+Q=\frac{\pi }{6}$, both $P$ and $Q$ are small, making $\sin P$ small, which won't satisfy (1).Thus, $P+Q=\frac{5\pi }{6}$.In a triangle, $P+Q+R=\pi$.$R=\pi -(P+Q)=\pi -\frac{5\pi }{6}=\frac{\pi }{6}$.