5. Determine the angle in the triangle that is $20 \%$ smaller than the angle between the bisectors of the other two angles of the triangle.
Solution. Let the angle be $\alpha$, and let $\varphi$ denote the angle between the bisectors of the other two angles $\beta$ and $\gamma$ of the triangle (see diagram). Then, from the condition, we have

![](https://cdn.mathpix.com/cropped/2024_06_05_dc4d34c0a9dd2a3ba572g-07.jpg?height=291&width=726&top_left_y=936&top_left_x=740)
that

$$
\alpha=80 \% \varphi=\frac{80}{100} \varphi=\frac{4}{5} \varphi
$$

From the property that the sum of the angles in a triangle is $180^{\circ}$, we have the following equalities

$$
\alpha+\beta+\gamma=180^{\circ} \text { and } \varphi+\frac{1}{2} \beta+\frac{1}{2} \gamma=180^{\circ}
$$

By substituting $\alpha=\frac{4}{5} \varphi$ into the first equation and rearranging the terms, we get

$$
\left(\frac{4}{5} \varphi+\frac{1}{2} \beta+\frac{1}{2} \gamma\right)+\left(\frac{1}{2} \beta+\frac{1}{2} \gamma\right)=180^{\circ}
$$

and by rearranging the terms in the second equation, we get

$$
\left(\frac{4}{5} \varphi+\frac{1}{2} \beta+\frac{1}{2} \gamma\right)+\frac{1}{5} \varphi=180^{\circ}
$$

From the last two equations, it is clear that $\frac{1}{2} \beta+\frac{1}{2} \gamma=\frac{1}{5} \varphi$. On the other hand, $\frac{1}{2} \beta+\frac{1}{2} \gamma=180^{\circ}-\varphi$. It follows that $\frac{1}{5} \varphi=180^{\circ}-\varphi$, from which $\frac{6}{5} \varphi=180^{\circ}$, and finally $\varphi=180^{\circ} \cdot \frac{5}{6}=150^{\circ}$. Then, the desired angle in the triangle is $\alpha=\frac{4}{5} \varphi=\frac{4}{5} \cdot 150^{\circ}=120^{\circ}$.

## VII Section