Task B-2.2. Points $H$ and $N$ are the feet of the altitudes from vertex $A$ and vertex $B$ of an acute-angled triangle $ABC$. The length of the altitude from vertex $A$ is $5 \sqrt{3} \mathrm{~cm}$, the length of side $\overline{A B}$ is $14 \mathrm{~cm}$, and the measure of the angle between the altitudes $\overline{A H}$ and $\overline{B N}$ is $60^{\circ}$. Determine the lengths of the remaining sides of the triangle and the length of segment $\overline{H N}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6bd7d8198e5de61c26bbg-07.jpg?height=716&width=770&top_left_y=1355&top_left_x=648)

Since the given triangle is acute-angled, we conclude that the measure of angle $\angle B O H$ is $60^{\circ}$, so the measure of angle $\angle N B C = 90^{\circ} - 60^{\circ} = 30^{\circ}$, and the measure of angle $\angle A C B = 60^{\circ}$. From triangle $A H C$, we get $|A C| = \frac{|A H|}{\sin 60^{\circ}} = 10 \text{ cm}$.

Then, $|C H| = \frac{1}{2}|A C| = 5 \text{ cm}$.

From triangle $A B H$, we get

$$
|B H|^{2} = |A B|^{2} - |A H|^{2} = 196 - (5 \sqrt{3})^{2} = 121
$$

so $|B H| = 11 \text{ cm}$, and thus $|B C| = 11 + 5 = 16 \text{ cm}$.

It remains to determine the length of segment $\overline{H N}$.

From triangle $B N C$, we get $|N C| = |B C| \cos 60^{\circ} = 16 \cdot \frac{1}{2} = 8 \text{ cm}$. Triangles $A B C$ and $H N C$ are similar because they share an angle and the ratio of their sides is $\frac{|H C|}{|A C|} = \frac{|N C|}{|B C|}$, i.e., $\frac{5}{10} = \frac{8}{16} = \frac{1}{2}$.

We conclude that the ratio of the sides $\frac{|H N|}{|A B|} = \frac{1}{2}$, so the length of segment $\overline{H N}$ is $\frac{1}{2} \cdot 14 = 7 \text{ cm}$.

The desired length can also be calculated using the height $v$ from vertex $H$ in triangle $H N C$.

$$
\begin{aligned}
& v = |H C| \cdot \sin 60^{\circ} = 5 \frac{\sqrt{3}}{2} \text{ cm} \\
& |C D| = \frac{1}{2}|H C| = \frac{5}{2} \text{ cm} \\
& |D N| = |C N| - |C D| = 5.5 \text{ cm} \\
& |H N|^{2} = v^{2} + |D N|^{2} \Rightarrow |H N| = 7 \text{ cm}
\end{aligned}
$$