Problem 8

Given that in triangle $\triangle ABC$, $AB=AC$, we can deduce that $\triangle ABC$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, we have:

1. $\angle B = \angle C$

Given that $\angle A = 80^{\circ}$, we know that the sum of angles in any triangle is $180^{\circ}$. Therefore, we can calculate $\angle B$ (and $\angle C$) as follows:

2. $\angle A + \angle B + \angle C = 180^{\circ}$

Substituting the given and derived values:

3. $80^{\circ} + \angle B + \angle B = 180^{\circ}$

Since $\angle B = \angle C$, we have:

4. $80^{\circ} + 2\angle B = 180^{\circ}$

Solving for $\angle B$:

5. $2\angle B = 180^{\circ} - 80^{\circ}$

6. $2\angle B = 100^{\circ}$

7. $\angle B = \frac{100^{\circ}}{2}$

8. $\angle B = 50^{\circ}$

Therefore, the correct answer is $\boxed{C: 50^{\circ}}$.