Macedonian Mathematical Competitions

Let us denote with $\alpha, \beta, \gamma$ the angles in the triangle $ABC$ at the vertices $A, B, C$ correspondently and with $\alpha_1, \beta_1, \gamma_1$ the correspondent exterior angles. Let $D$ be the base of the altitude from $C$, $E$ be the base of the bisector of the angle $ACB$ and $F$ be the intersection of the bisectors $AY$ and $BN$ of the exterior angles at the vertices $A$ and $B$ correspondently. Then $\angle DCE = 90^\circ$, $\angle AFB = 61^\circ$, $\angle XAY = \angle YAC = \frac{\alpha_1}{2}$, $\angle CBN = \angle NBM = \frac{\beta_1}{2}$ and $\angle ACE = \angle ECB = \frac{\gamma_1}{2}$.

For the angles in the triangle $ABF$ we have that

$\angle FAB = \angle XAY = \frac{\alpha_1}{2}$ and $\angle ABF = \angle NBM = \frac{\beta_1}{2}$ as opposite angles. Now we obtain $\frac{\alpha_1}{2} + \frac{\beta_1}{2} + 61^\circ = 180^\circ$ (because the sum of the angles in every triangle is $180^\circ$). Hence $\alpha_1 + \beta_1 = 238^\circ$. Because $\alpha_1 = 180^\circ - \alpha$ and $\beta_1 = 180^\circ - \beta$ if substitute in the previous equality we get $180^\circ - \alpha + 180^\circ - \beta = 238^\circ$. Hence $\alpha + \beta = 122^\circ$. $\gamma = 180^\circ - (\alpha + \beta) = 180^\circ - 122^\circ = 58^\circ$.

From the right-angled triangle $ADC$ we have
$$\alpha = 90^\circ - \angle ACD = 90^\circ - (\frac{\gamma}{2} - \angle CDE) = 90^\circ - (\frac{58^\circ}{2} - 9^\circ) = 70^\circ.$$
Now we get that $\beta = 122^\circ - \alpha = 122^\circ - 70^\circ = 52^\circ$.