## Task B-1.5.

The internal angles of a triangle are in the ratio 2 : 3 : 7. Determine the length of the longest side of the triangle, if the shortest side has a length of $1 \mathrm{~cm}$.
## Solution.

From $\alpha: \beta: \gamma=2: 3: 7$ and $\alpha+\beta+\gamma=180^{\circ}$, we get $\alpha=30^{\circ}, \beta=45^{\circ}$, and $\gamma=105^{\circ}$. 1 point

Thus, we have the following sketch.

![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-05.jpg?height=800&width=1036&top_left_y=642&top_left_x=430)

Since $\alpha$ is the smallest and $\gamma$ is the largest angle, we conclude that the shortest side is $\overline{B C}$, and the longest side is $\overline{A B}$.

Since $\overline{B C}$ is the diagonal of the square $B F C D$ and its length is $1 \mathrm{~cm}$, it follows that $|B D|=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \mathrm{~cm}$.

Triangle $A E C$ is equilateral with side length $2|C D|=\sqrt{2}$.

The length of $\overline{A D}$ is the height of the equilateral triangle $A E C$ with side length $\sqrt{2}$, so $|A D|=\frac{\sqrt{2} \cdot \sqrt{3}}{2}=\frac{\sqrt{6}}{2}$.

Therefore, the length of the longest side $|A B|=|A D|+|D B|=\frac{1}{2}(\sqrt{6}+\sqrt{2}) \mathrm{~cm}$.