1. [3] Triangle $A B C$ is isosceles, and $\angle A B C=x^{\circ}$. If the sum of the possible measures of $\angle B A C$ is $240^{\circ}$, find $x$.
Answer: $\square$
There are three possible triangles: either $\angle A B C=\angle B C A$, in which case $\angle B A C=180-2 x, \angle A B C=$ $\angle B A C$, in which case $\angle B A C=x$, or $\angle B A C=\angle B C A$, in which case $\angle B A C=\frac{180-x}{2}$. These sum to $\frac{540-3 x}{2}$, so we have $\frac{540-3 x}{2}=240 \Longrightarrow x=20$.