Geometry
# Solving Triangles

If $CD=\sqrt{3},$ $PQ=2,$ and $\angle PDQ=30^\circ,$ find $CP.$

In triangle $ABC$, $\angle BAC = 2 \angle ABC$, $AB= 31, AC = 13$. What is $BC^2$?

If $AP = AC + 2\cdot CB$, what is the measure $\angle ABP$ in degrees?