Hey, good to see you again. Step 1: Find the third angle in triangle ABC. 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 to find BC. The Sine Rule states that for a triangle with sides a, b, c and opposite angles A, B, C: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$ In triangle ABC, we want to find side BC (let's call it $a$), and we know angle CAB (A). We also know side AC (let's call it $b$) and angle ABC (B). $$ \frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)} $$ Substitute the given values: $$ \frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)} $$ Solve for BC: $$ BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)} $$ $$ BC = \frac{50 \times 0.7071}{0.9848} $$ $$ BC = \frac{35.355}{0.9848} $$ $$ BC \approx 35.90 \text{ units} $$ The length of BC is approximately $\boxed{35.90}$. Send me the next one 📸