Here are the solutions to your trigonometry problems: 1. In the triangle ABC, ABC = 80, CAB = 45 and AC = 50. Find BC. Step 1: Find the third angle of the triangle. The sum of angles in a triangle is $180^\circ$. $$ \angle ACB = 180^\circ - \angle ABC - \angle CAB $$ $$ \angle ACB = 180^\circ - 80^\circ - 45^\circ $$ $$ \angle ACB = 55^\circ $$ Step 2: Apply the Sine Rule. The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. We want to find BC (side $a$). $$ \frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)} $$ $$ \frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)} $$ Step 3: Solve for BC. $$ BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)} $$ $$ BC = \frac{50 \times 0.7071}{0.9848} $$ $$ BC = \frac{35.355}{0.9848} $$ $$ BC \approx 35.900 $$ The length of BC is approximately $\boxed{\text{35.90 units}}$. 2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 6.5. Find PQ. Step 1: Find the third angle of the triangle. $$ \angle PQR = 180^\circ - \angle RPQ - \angle QRP $$ $$ \angle PQR = 180^\circ - 60^\circ - 50^\circ $$ $$ \angle PQR = 70^\circ $$ Step 2: Apply the Sine Rule. We want to find PQ (side $r$). $$ \frac{PQ}{\sin(\angle QRP)} = \frac{RP}{\sin(\angle PQR)} $$ $$ \frac{PQ}{\sin(50^\circ)} = \frac{6.5}{\sin(70^\circ)} $$ Step 3: Solve for PQ. $$ PQ = \frac{6.5 \times \sin(50^\circ)}{\sin(70^\circ)} $$ $$ PQ = \frac{6.5 \times 0.7660}{0.9397} $$ $$ PQ = \frac{4.979}{0.9397} $$ $$ PQ \approx 5.298 $$ The length of PQ is approximately $\boxed{\text{5.30 units}}$. 3 done, 2 left today. You're making progress.