In triangle $ABC$, the angle bisectors of angles $\alpha$ and $\gamma$ intersect at point $M$. Calculate angle $\beta$ if it is known that $\beta$ is equal to half of angle $AMC$.

1. Problem Description

In triangle $ABC$, the angle bisectors of angles $\alpha$ and $\gamma$ intersect at point $M$. Calculate angle $\beta$ if it is known that $\beta$ is equal to half of angle $AMC$.

2. Solution Steps

Let $\alpha$, $\beta$, and $\gamma$ be the angles of triangle $ABC$. We are given that the angle bisectors of $\alpha$ and $\gamma$ intersect at point $M$.

We know that $\alpha + \beta + \gamma = 180^\circ$.

The angle bisectors divide $\alpha$ and $\gamma$ into $\frac{\alpha}{2}$ and $\frac{\gamma}{2}$, respectively.

In triangle $AMC$, the sum of the angles is $180^\circ$, so

$\angle MAC + \angle MCA + \angle AMC = 180^\circ$.

Substituting the bisected angles, we have

$\frac{\alpha}{2} + \frac{\gamma}{2} + \angle AMC = 180^\circ$.

Thus, $\angle AMC = 180^\circ - \frac{\alpha}{2} - \frac{\gamma}{2}$.

We are given that $\beta = \frac{1}{2} \angle AMC$. Therefore,

$\beta = \frac{1}{2} (180^\circ - \frac{\alpha}{2} - \frac{\gamma}{2}) = 90^\circ - \frac{\alpha}{4} - \frac{\gamma}{4}$.

Since $\alpha + \beta + \gamma = 180^\circ$, we have $\alpha + \gamma = 180^\circ - \beta$.

So, $\frac{\alpha}{4} + \frac{\gamma}{4} = \frac{1}{4}(\alpha + \gamma) = \frac{1}{4}(180^\circ - \beta) = 45^\circ - \frac{\beta}{4}$.

Substituting this into the equation for $\beta$:

$\beta = 90^\circ - (45^\circ - \frac{\beta}{4}) = 90^\circ - 45^\circ + \frac{\beta}{4} = 45^\circ + \frac{\beta}{4}$.

Multiplying by 4, we get $4\beta = 180^\circ + \beta$.

Subtracting $\beta$ from both sides, we have $3\beta = 180^\circ$.

Therefore, $\beta = \frac{180^\circ}{3} = 60^\circ$.

3. Final Answer

The angle $\beta$ is 60 degrees.