 Solution

Call the centre of the circle $O$, the left endpoint on the diameter $A$, the point of tangency $T$, and the external intersection $E$. Since $OA=OT$ are radii, the isosceles triangle $AOT$ has equal base angles: $\angle OAT=\angle ATO=30^{\circ }$. Therefore $\angle AOT=120^{\circ }$, and, because $OA$ and $OE$ point in opposite directions, $\angle TOE=60^{\circ }$. The tangent to a circle $TE$ is perpendicular to the radius $OT$, so in the right triangle $OTE$,