2B. In an acute triangle $ABC$, the altitude from vertex $C$ divides the opposite side into two segments $\overline{AD}$ and $\overline{DB}$ with lengths $\overline{AD}=3 \mathrm{~cm}$ and $\overline{DB}=2 \mathrm{~cm}$. The angle at vertex $A$ measures $60^{\circ}$. Determine the lengths of the sides of triangle $ABC$ as well as the length of the altitude of triangle $ABC$ dropped from vertex $A$.
Solution. Clearly, $\overline{A B}=5 \mathrm{~cm}$. Since the angle at vertex $A$ is $60^{\circ}$ and triangle $A D C$ is

![](https://cdn.mathpix.com/cropped/2024_06_05_6152dee85d0865b6f018g-1.jpg?height=488&width=672&top_left_y=1674&top_left_x=844)
a right triangle, we have that angle $\angle D C A$ is $30^{\circ}$, from which it follows that $\overline{A C}=2 \overline{A D}=6 \mathrm{~cm}$. Using the Pythagorean theorem in triangle $A D C$, we get

$$
\overline{D C}=\sqrt{\overline{A C}^{2}-\overline{A D}^{2}}=\sqrt{6^{2}-3^{2}}=5 \mathrm{~cm}
$$

Using the Pythagorean theorem in triangle $B D C$, we get

$$
\overline{B C}=\sqrt{\overline{D C}^{2}+\overline{B D}^{2}}=\sqrt{5^{2}+4^{2}}=\sqrt{29} \mathrm{~cm}
$$

Next, for the area of triangle $A B C$, we have $P_{A B C}=\frac{\overline{A B} \cdot \overline{C D}}{2}=\frac{5 \cdot 5}{2}=\frac{25}{2}$. On the other hand, $P_{A B C}=\frac{\overline{B C} \cdot \overline{A E}}{2}$, so $\frac{25}{2}=\frac{\sqrt{29} \cdot \overline{A E}}{2}$. Finally, the height from vertex $A$ is $\overline{A E}=\frac{25}{\sqrt{29}} \mathrm{~cm}$.