Task B-1.4. In an acute-angled triangle $ABC$, the height from vertex $C$ divides the opposite side into two segments $\overline{AD}$ and $\overline{DB}$ with lengths $|\overline{AD}|=4 \text{~cm}$ and $|\overline{DB}|=1 \text{~cm}$. The measure of the angle at vertex $A$ is $60^{\circ}$. Determine the lengths of the sides of triangle $ABC$ and the length of the height from vertex $A$ of this triangle.
## Solution.

Obviously, the length of side $\overline{A B}$ is $5 \mathrm{~cm}$.

In the right triangle $A D C$, the remaining angles are $60^{\circ}$ and $30^{\circ}$, so this triangle is half of an equilateral triangle whose side length is equal to $|\overline{A C}|=2 \cdot|\overline{A D}|=8 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-03.jpg?height=608&width=458&top_left_y=417&top_left_x=801)

(2 points)

The height of triangle $A B C$ is also the height of the equilateral triangle, so $v=\frac{8 \sqrt{3}}{2}=4 \sqrt{3} \mathrm{~cm}$.

Then, $|\overline{B C}|=\sqrt{v^{2}+1}=7 \mathrm{~cm}$.

The area of the given triangle is $\frac{5 \cdot 4 \sqrt{3}}{2}=10 \sqrt{3} \mathrm{~cm}^{2}$. The area can also be expressed as

$$
P=\frac{|\overline{B C}| \cdot v_{a}}{2}
$$

By substituting into the formula, we calculate the height from vertex $A$

$$
\begin{aligned}
& 10 \sqrt{3}=\frac{7 \cdot v_{a}}{2} \\
& v_{a}=\frac{20 \sqrt{3}}{7} \mathrm{~cm}
\end{aligned}
$$