In a ΔPQR, if 3sinP+4cosQ=16 and 4sinQ+3cosP=1, Then the angle R is equal to
π/4
3π/4
5π/6
π/6
Squaring and adding the given relations we get
16+9+24sin(P+Q)=37⇒sin(P+Q)=1/2⇒sinR=1/2
⇒R=π/6 or 5π/6
If R=5π/6, then P<π/6⇒3sinP<3/2.
⇒3sinP+4cosQ<3/2+4<6
So R≠5π/6