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_dd45b8ae4a4e0390a616g-1.jpg?height=260&width=374&top_left_y=2076&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 70 , then $t=70$, and so

$$
\angle A B C=180^{\circ}-2 \cdot 70^{\circ}=180^{\circ}-140^{\circ}=40^{\circ}
$$

ANSWER: $5,70,40^{\circ}$