## Solution.

Let $a, b$ and $c$ be the lengths of the sides $\overline{B C}, \overline{A C}$ and $\overline{A B}$ of triangle $A B C$, respectively, and let $\alpha$ be the measure of the angle at vertex $A$.

Using the formula for the area of a triangle involving the sine of an angle gives

$$
5=P(A B C)=\frac{1}{2} b c \sin \alpha
$$

from which it follows that $\sin \alpha=\frac{10}{b c}$.

Applying the cosine rule to triangle $A B C$ yields

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha
$$

from which it follows that

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{17}{2 b c}
$$

where in the last equality we used the condition of the problem $b^{2}+c^{2}=17+a^{2}$.

Finally,

$$
\operatorname{tg} \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{10}{b c} \cdot \frac{2 b c}{17}=\frac{20}{17}
$$
## Task A-3.3.

Given is a triangle $A B C$ with an area of 5. If for the lengths of the sides of this triangle the equality $|A B|^{2}+|A C|^{2}=17+|B C|^{2}$ holds, determine the tangent of the angle $\varangle C A B$.