## Task B-2.2.

In triangle $ABC$, the measures of the angles at vertex $A$ and vertex $C$ are $\alpha=60^{\circ}$ and $\gamma=75^{\circ}$, respectively. Calculate the distance from the orthocenter of triangle $ABC$ to vertex $B$ if $|BC|=8\sqrt{6}$.
## First Solution.

Let $H$ be the orthocenter of triangle $ABC$, and points $D, E$, and $F$ be the feet of the altitudes from vertices $C, A$, and $B$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-09.jpg?height=794&width=1151&top_left_y=648&top_left_x=384)

Then the measure of angle $\varangle ABC$ is $180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}$.

(scoring the above diagram with marked angles, altitudes, and orthocenter)

From triangle $BCD$, we have

$$
\sin 45^{\circ} = \frac{|CD|}{8 \sqrt{6}}
$$

or

$$
|CD| = \frac{\sqrt{2}}{2} \cdot 8 \sqrt{6} = 8 \sqrt{3}
$$

From triangle $ACD$, we get $\operatorname{tg} 60^{\circ} = \frac{|CD|}{|AD|}$, or

$$
|AD| = \frac{8 \sqrt{3}}{\sqrt{3}} = 8
$$

and thus $|HD| = |AD| = 8$. 1 point

From triangle $BHD$, we get $\sin 30^{\circ} = \frac{|HD|}{|BH|}$, or $|BH| = \frac{8}{\frac{1}{2}} = 16$.