Problem — Tangents at sixty degrees

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From a point $A$ outside a circle with centre $O$ and radius $r = 5$, draw the two tangents, which touch the circle at $B$ and $C$ and form an angle $\widehat{BAC} = 60^\circ$ between them. Compute the distance $\overline{OA}$ and the length of the tangent segments, and compare $\overline{AO}$, $\overline{AB}$, $\overline{AC}$.

Solution

The line $AO$ is the bisector of the angle formed by the two tangents, so $\widehat{BAO} = 30^\circ$. Triangle $OBA$ is right-angled at $B$, hence $\sin\widehat{BAO} = \frac{\overline{OB}}{\overline{OA}} \implies \overline{OA} = \frac{r}{\sin 30^\circ} = \frac{5}{1/2} = 10.$ The tangent segment is $\overline{AB} = \sqrt{\overline{OA}^{\,2} - r^2} = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3} \approx 8.66,$ and $\overline{AC} = \overline{AB}$ (tangent segments from the same point). Comparing the three segments, $\overline{AO} = 10 > \overline{AB} = \overline{AC} = 5\sqrt{3} \approx 8.66$. $\boxed{\overline{OA} = 10, \qquad \overline{AB} = \overline{AC} = 5\sqrt{3} \approx 8.66}$