Let $t$ be TNYWR.

In the diagram, $\triangle A B C$ is isosceles with $A B=B C$. If $\angle B A C=t^{\circ}$, what is the measure of $\angle A B C$, in degrees?

![](https://cdn.mathpix.com/cropped/2024_04_17_23a04f66c1167e793e90g-1.jpg?height=263&width=374&top_left_y=2053&top_left_x=903)
Since $A B=B C$, then $\angle B C A=\angle B A C=t^{\circ}$.

Therefore, $\angle A B C=180^{\circ}-\angle B C A-\angle B A C=180^{\circ}-2 t^{\circ}$.

Since the answer to (b) is 52 , then $t=52$, and so

$$
\angle A B C=180^{\circ}-2 \cdot 52^{\circ}=180^{\circ}-104^{\circ}=76^{\circ}
$$

ANSWER: $4,52,76^{\circ}$