In a $∆ \mathrm{PQR}$ if $3\mathrm{sinP} + 4\mathrm{cosQ} = 6$ and $4\sin \mathrm{Q}+3 \cos \mathrm{P}=1$ then the acute angle $\mathrm{R}$ is $\dots \dots$

$$\begin{gathered} 3\sin \mathrm{P}+4\cos \mathrm{Q}=6 -\left(1\right) \\ 4\sin \mathrm{Q}+3\cos \mathrm{P}=1 -\left(2\right) \\ {\left(1\right)}^2 + {\left(2\right)}^2 \Rightarrow \\ 9{\sin }^2\mathrm{P}+16{\cos }^2\mathrm{Q}+24\mathrm{sinP} \cos \mathrm{Q}+16{\sin }^2\mathrm{Q}+9{\cos }^2\mathrm{P}+24\cos \mathrm{P} \sin \mathrm{Q}=36+1 \\ 9({\sin }^2\mathrm{P}+{\cos }^2\mathrm{P})+16({\sin }^2\mathrm{Q}+{\cos }^2\mathrm{Q})+24(\sin \mathrm{P} \cos \mathrm{Q}+\cos \mathrm{P} \sin \mathrm{Q})=37 \\ 9+16+24 \sin (\mathrm{P}+\mathrm{Q})=37 \\ 24 \sin (\mathrm{P}+\mathrm{Q})=12 \\ \sin (\mathrm{P}+\mathrm{Q})=\frac{1}{2} \\ \mathrm{P}+\mathrm{Q}=\frac{\mathrm{\pi }}{6} \\ \angle R=\mathrm{\pi }-\frac{\mathrm{\pi }}{6}=\frac{5\mathrm{\pi }}{6} \end{gathered}$$