[ Concurrency of heights. Angles between heights. [Trigonometric ratios in a right-angled triangle ]

In triangle $A B C$, angle $\angle B$ is obtuse, the extensions of heights $A M$ and $C N$ intersect at point $O, \angle B A C=\alpha, \angle B C A=\gamma, A C=b$. Find the distance from point $O$ to the line $A C$.
The height $O H$ of the acute triangle $A O C$ passes through the point $B$.

## Solution

Let $H$ be the projection of point $O$ onto $A C$. Since $A N, C M$, and $O H$ are the altitudes of the acute triangle $A O C$, the line $O H$ passes through point $B$ (the altitudes of a triangle intersect at one point), and point $H$ lies on the segment $A C$, $\angle A O H = \angle A C M = \gamma$, $\angle C O H = \angle C A N = \alpha$.

Let $O H = x$. From the right triangles $A H O$ and $C H O$, we find that

$$
\mathrm{AH} = \mathrm{OH} \cdot \operatorname{tg} \angle \mathrm{AOH} = x \operatorname{tg} \gamma, \mathrm{CH} = \mathrm{OH} \cdot \operatorname{tg} \angle \mathrm{COH} = x \operatorname{tg} \alpha
$$

Since $A H + H B = A C$, we have the equation

$$
x \operatorname{tg} \gamma + x \operatorname{tg} \alpha = b
$$

From which we find that

$$
O H = x = \frac{b}{\operatorname{tg} \gamma + \operatorname{tg} \alpha} = \frac{b \cos \alpha \cos \gamma}{\sin (\alpha + \gamma)}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-29.jpg?height=466&width=671&top_left_y=956&top_left_x=705)

## Answer

$b \cdot \frac{\cos \alpha \cos \gamma}{\sin (\alpha + \gamma)}$.