[tex]\frac{\sin(R)}{PQ} = \frac{\sin(Q)}{PR}[/tex]
[tex]\frac{\sin(R)}{24.7} = \frac{\sin(43^{\circ})}{16.9}[/tex]
[tex]\sin(R) = 24.7*\frac{\sin(43^{\circ})}{16.9}[/tex]
[tex]\sin(R) \approx 0.9967668[/tex]
[tex]R \approx \arcsin(0.9967668)[/tex]
[tex]R \approx 85.39137896^{\circ}[/tex]
[tex]R \approx 85^{\circ}[/tex]
The final answer is 85 degrees
