Answer to Question #117464 in Trigonometry for Cecilia LeBrun

Question

Solve for each triangle. Round answers to nearest tenth. Triangle PQR angle P= 39, r=31 and p= 29

Answer 1

$\frac{RQ}{\sin P}=\frac{PQ}{\sin R}=\frac{PR}{\sin Q};\\\frac{RQ}{\sin P}=\frac{29}{\sin 39}=\frac{29}{0.485}=59.794;\\\frac{PQ}{\sin R}=\frac{31}{\sin R}=59.794;\sin R=\frac{31}{59.794};R=31^{\circ}.198;\\180^{\circ}-39^{\circ}-31^{\circ}.198=109^{\circ}.802;\\Q=109^{\circ}.802;\sin109^{\circ}.802=0.941;\\\frac{PR}{\sin Q}=59.794;PR=59.794\times0.941=56.266;\\Answer:PR=59.8;Q=109^\circ.8;R=31^\circ.2$