319. The vertices of the triangle are given by the radius vectors $\bar{a}=$ $=\{1 ; 2\}, \bar{b}=\{2 ; 5\}, \bar{c}=\{3 ; 4\}$. Determine the sides and internal angles of the triangle.
Solution. From the condition of the problem, the vertices of the triangle are $A(1 ; 2), B(2 ; 5), C(3 ; 4)$.

Therefore, the vectors coinciding with the sides of the triangle are: $\overline{A B}=\{1 ; 3\}, \overline{A C}=\{2 ; 2\}, \overline{B C}=\{1 ;-1\}$ (Fig. 45). Let's find the lengths of the sides of the triangle: $A B=|\overline{A B}|=\sqrt{1+9}=$ $=\sqrt{10}, A C=|\overline{A C}|=\sqrt{4+4}=2 \sqrt{2}, B C=|\overline{B C}|=\sqrt{1+1}=\sqrt{2}$.

It is not difficult to see that

$$
A C^{2}+B C^{2}=A B^{2}
$$

Therefore, triangle $A B C$ is a right triangle, with the hypotenuse being side $A B$, and angle $C$ is a right angle. To determine the cosine of angle $A$, we use formula (16):

$$
\cos A=\frac{\overline{A B} \cdot \overline{A C}}{|\overline{A B}| \cdot|\overline{A C}|}=\frac{1 \cdot 2+3 \cdot 2}{\sqrt{10} \cdot 2 \sqrt{2}}=\frac{2 \sqrt{5}}{5} \approx 0.8996
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-074.jpg?height=310&width=341&top_left_y=1082&top_left_x=70)

Fig. 45

Using trigonometric function tables, we find $A \approx 25^{\circ} 54^{\prime}$.

Since the triangle is a right triangle, angle $B=90^{\circ}-A \approx 90^{\circ}-25^{\circ} 54^{\prime}=$ $=64^{\circ} 6^{\prime}$.