## Task B-4.5.

In triangle $A B C$, the altitude from vertex $A$ divides the opposite side into segments of length $2 \mathrm{~cm}$ and $10 \mathrm{~cm}$, and $\operatorname{tg} \varangle C A B=\frac{9}{2}$. Calculate the area of triangle $A B C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_a25c851e6461b5bf7f1cg-18.jpg?height=337&width=597&top_left_y=1822&top_left_x=661)

Let $\alpha$ and $\beta$ be the parts into which the height from vertex $A$ divides angle $CAB$.

We have

$$
\begin{array}{cl}
\operatorname{tg} \alpha=\frac{2}{v}, \quad \operatorname{tg} \beta=\frac{10}{v} \\
\text{and} \quad & \operatorname{tg}(\alpha+\beta)=\frac{9}{2}
\end{array}
$$

Using the formula for the tangent of a sum, we get

$$
\begin{array}{r}
\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=\frac{9}{2} \\
\frac{\frac{2}{v}+\frac{10}{v}}{1-\frac{2}{v} \cdot \frac{10}{v}}=\frac{12 v}{v^{2}-20}=\frac{9}{2} \\
9 v^{2}-24 v-180=0 \\
3 v^{2}-8 v-60=0
\end{array}
$$

1 point

The positive solution to this equation $v=6$ is the height of triangle $ABC$.

1 point

The desired area is then

$$
P=\frac{(2+10) \cdot 6}{2}=36 \mathrm{~cm}^{2}
$$