In a triangle PQR, if $3\sin(P)+4\cos(Q)=6$ and $4\sin(Q)+3cos(P)=1$ then what will be the angle $R$ ?
On squaring the equations and adding them and solving the resultant equation I find that $R=30^o$ or $R=150^o$. My book says that $R$ cannot be $150^o$. I cannot understand why! Any ideas?
If R is 150 deg, then sin P is at most 1/2 and this forces 4cos Q to be at least 4.5, which is not possible.