2.1. (2 points) On the side $B C$ of triangle $A B C$, a point $D$ is chosen such that $\angle B A D=50^{\circ}$, $\angle C A D=20^{\circ}$, and $A D=B D$. Find $\cos \angle C$.
Answer: $\frac{\sqrt{3}}{2} \approx 0.87$.

Solution. Triangle $A B D$ is isosceles, $\angle A D B=\angle B A D=50^{\circ}$, then $\angle A B D=80^{\circ}$, $\angle B A C=70^{\circ}$, therefore $\angle C=180^{\circ}-70^{\circ}-80^{\circ}=30^{\circ}, \cos \angle C=\frac{\sqrt{3}}{2} \approx 0.87$.