Find the third angle in triangle ABC.

The sum of angles in a triangle is $180^\circ$.

$$\angle ACB = 180^\circ - \angle ABC - \angle CAB$$ $$\angle ACB = 180^\circ - 80^\circ - 45^\circ$$ $$\angle ACB = 55^\circ$$

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 35.90.