Two tangents are drawn from a point $A$ to a circle with center $O$, touching it at $B$ and $C$. Let $H$ be the orthocenter of triangle $A B C$, given that $\angle B A C=40^{\circ}$, find the value of the angle $\angle H C O$.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-03.jpg?height=368&width=631&top_left_y=2163&top_left_x=815)
Solution

Since $AC$ is tangent to the circle, we have $\angle A C O=90^{\circ}$. Thus

$$
\begin{aligned}
\angle H C O & =90^{\circ}-\angle A C F \\
& =\angle C A F \\
& =40^{\circ}
\end{aligned}
$$