Problem 5. Calculate the angles in triangle $ABC$, if the angle between the height drawn from vertex $C$ and the bisector of angle $ACB$ is $9^{\circ}$, and the angle between the bisectors of the exterior angles at vertices $A$ and $B$ is $61^{\circ}$.
Solution. Let's denote the angles of triangle $ABC$ at vertices $A, B, C$ as $\alpha, \beta, \gamma$ respectively, and the corresponding exterior angles as $\alpha_{1}, \beta_{1}, \gamma_{1}$. Let $D$ be the foot of the altitude from vertex $C$, point $E$ be the foot of the angle bisector of $\angle ACB$, and let $F$ be the intersection of the angle bisectors $AY$ and $BN$ of the exterior angles at vertices $A$ and $B$ respectively (see diagram).

![](https://cdn.mathpix.com/cropped/2024_06_05_028d7bfd550280da1c84g-06.jpg?height=503&width=510&top_left_y=1378&top_left_x=954)

From the given conditions,

$$
\begin{gathered}
\measuredangle DCE = 9^{\circ}, \measuredangle AFB = 61^{\circ}, \measuredangle XAY = \measuredangle YAC = \frac{\alpha_{1}}{2} \\
\measuredangle CBN = \measuredangle NBM = \frac{\beta_{1}}{2} \text{ and } \measuredangle ACE = \measuredangle ECB = \frac{\gamma}{2}.
\end{gathered}
$$

For the angles in triangle $ABF$, we have

$$
\measuredangle FAB = \measuredangle XAY = \frac{\alpha_{1}}{2} \text{ and } \measuredangle ABF = \measuredangle NBM = \frac{\beta_{1}}{2}
$$

as alternate interior angles, so from the sum of angles in a triangle we get

$$
\frac{\alpha_{1}}{2} + \frac{\beta_{1}}{2} + 61^{\circ} = 180^{\circ}
$$

from which it follows that

$$
\alpha_{1} + \beta_{1} = 238^{\circ}. \text{ Since } \alpha_{1} = 180^{\circ} - \alpha \text{ and } \beta_{1} = 180^{\circ} - \beta
$$

substituting into the last equation gives

$$
180^{\circ} - \alpha + 180^{\circ} - \beta = 238^{\circ}
$$

from which $\alpha + \beta = 122^{\circ}$, so

$$
\gamma = 180^{\circ} - (\alpha + \beta) = 180^{\circ} - 122^{\circ} = 58^{\circ}
$$

Then, from the right triangle $ADC$, we have

$$
\alpha = 90^{\circ} - \measuredangle ACD = 90^{\circ} - \left(\frac{\gamma}{2} - \measuredangle CDE\right) = 90^{\circ} - \left(\frac{58^{\circ}}{2} - 9^{\circ}\right) = 70^{\circ}
$$

and finally

$$
\beta = 122^{\circ} - \alpha = 122^{\circ} - 70^{\circ} = 52^{\circ}
$$

## VII Section