Here's the solution for question 1. 1) In the triangle ABC, $\angle ABC = 80^\circ$, $\angle CAB = 45^\circ$ and $AC = 50$. Find BC. Step 1: Find the third angle, $\angle BCA$. The sum of angles in a triangle is $180^\circ$. $$ \angle BCA = 180^\circ - \angle ABC - \angle CAB $$ $$ \angle BCA = 180^\circ - 80^\circ - 45^\circ $$ $$ \angle BCA = 55^\circ $$ 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 side $BC$ (let's call it $a$), which is opposite $\angle CAB$ ($A$). We are given side $AC$ (let's call it $b$), which is opposite $\angle ABC$ ($B$). So, we use the relation: $$ \frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)} $$ Step 3: Substitute the given values and solve for $BC$. $$ \frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)} $$ $$ BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)} $$ $$ BC = \frac{50 \times 0.7071}{0.9848} $$ $$ BC = \frac{35.355}{0.9848} $$ $$ BC \approx 35.90 $$ The length of side BC is approximately $35.90$. $$ \boxed{\text{BC} \approx 35.90} $$ 3 done, 2 left today. You're making progress.