11.2. In an isosceles triangle \(ABC\) with base \(AC\), the angle between the bisector and the altitude drawn from vertex \(A\) is 60 degrees. Find the angles of triangle \(ABC\).
Answer: 20, 20, and 140 degrees.

Solution. Let $A P$ and $A M$ be the altitude and angle bisector, respectively, drawn from vertex $A$. Denote the measure of angle $BAC$ as $x$, then

$\angle PAC=90^{\circ}-x$ and $\angle MAC=\frac{x}{2}$.

1) Suppose first that point $M$ lies on segment $PC$, which occurs when $\angle MAC=\frac{x}{2} \leq \angle PAC=90^{\circ}-x$, that is, when $x \leq 60^{\circ}$. Then $\angle PAM=\angle PAC-\angle MAC=90^{\circ}-x-\frac{x}{2}=60^{\circ}$, from which $x=20$.
2) In the remaining case $x>60^{\circ}$, point $P$ lies on segment $MC$. Then $\angle PAM=\angle MAC-\angle PAC=\frac{x}{2}-90^{\circ}+x=60^{\circ}$, from which $x=100^{\circ}$, which is impossible for the base angle of an isosceles triangle.