2. Let $A B C$ be a right triangle with $\angle B C A=90^{\circ}$. A circle with diameter $A C$ intersects the hypotenuse $A B$ at $K$. If $B K: A K=1: 3$, find the measure of the angle $\angle B A C$.
Solution: The triangle $A K C$ is drawn in a semi-circle, so the angle $A K C$ is a right angle. Thus, $A K C$ and $A C B$ are similar triangles. WOLOG we assume that $A K=3$ and $B K=1$. Let $A C=x$. Then by similar triangles $\frac{3}{x}=\frac{x}{4}$, so $x=2 \sqrt{3}$. By the Pythagorean Theorem, $B C^{2}=4^{2}-12=4$ so $B C=2$. We see that $A B C$ is a $30-60-90$ triangle, with $B A C$ being the $30^{\circ}$ angle.