The problem states that two angles of a triangle are $60^\circ$ and $72^\circ$. We need to find the angles formed by the altitudes of the triangle originating from the vertices of the given angles.

1. Problem Description

The problem states that two angles of a triangle are $60^\circ$ and $72^\circ$. We need to find the angles formed by the altitudes of the triangle originating from the vertices of the given angles.

2. Solution Steps

Let the angles of the triangle be $A$, $B$, and $C$. We are given $A = 60^\circ$ and $B = 72^\circ$.

The sum of the angles in a triangle is $180^\circ$. Therefore,

$A + B + C = 180^\circ$

$60^\circ + 72^\circ + C = 180^\circ$

$132^\circ + C = 180^\circ$

$C = 180^\circ - 132^\circ$

$C = 48^\circ$

Let $h_A$, $h_B$, and $h_C$ be the altitudes originating from vertices $A$, $B$, and $C$ respectively.

Let the angles formed by the altitudes $h_A$ and $h_B$ be $\alpha$ and $\beta$. The point of intersection of the altitudes is the orthocenter.

The altitude from vertex $A$ is perpendicular to side $BC$, and the altitude from vertex $B$ is perpendicular to side $AC$. Therefore, the angle between $h_A$ and $h_B$ is supplementary to angle $C$.

$\alpha = 180^\circ - C$

$\alpha = 180^\circ - 48^\circ$

$\alpha = 132^\circ$

The other angle formed by $h_A$ and $h_B$ is

$\beta = 180^\circ - \alpha = 180^\circ - 132^\circ = 48^\circ$

Since the altitudes from $A$ and $B$ are perpendicular to the sides opposite to $A$ and $B$ respectively, they form right angles.

Let the altitudes $h_A$ and $h_B$ intersect at $H$. Consider the quadrilateral formed by vertices $A$, $B$, $H$ and the foot of the altitude from $A$ to $BC$, say $D$ and the foot of the altitude from $B$ to $AC$, say $E$. The quadrilateral is $CDHE$. The sum of angles in a quadrilateral is $360^\circ$. Thus, $\angle D + \angle E + \angle C + \angle DHE = 360^\circ$. We have $\angle D = \angle E = 90^\circ$ and $\angle C = 48^\circ$. So $90^\circ + 90^\circ + 48^\circ + \angle DHE = 360^\circ$, which implies $228^\circ + \angle DHE = 360^\circ$. Thus $\angle DHE = 360^\circ - 228^\circ = 132^\circ$. The angle formed is $132^\circ$, so the other angle is $180^\circ - 132^\circ = 48^\circ$.

The angles formed by $h_A$ and $h_C$ are supplementary to angle $B$.

$180^\circ - B = 180^\circ - 72^\circ = 108^\circ$

The other angle is $72^\circ$.

The angles formed by $h_B$ and $h_C$ are supplementary to angle $A$.

$180^\circ - A = 180^\circ - 60^\circ = 120^\circ$

The other angle is $60^\circ$.

3. Final Answer

The angles formed by the altitudes originating from the vertices of the given angles are $132^\circ$ and $48^\circ$. The other possible angles between the altitudes from vertices A and C are $108^\circ$ and $72^\circ$. The other possible angles between the altitudes from vertices B and C are $120^\circ$ and $60^\circ$. The problem asks for the angles formed by the altitudes originating from the vertices of the angles $60^\circ$ and $72^\circ$, which is the angle at the orthocenter, $132^\circ$. The supplementary angle to $C = 48^\circ$ is $132^\circ$, thus the angle is $132^\circ$. Also, it can be $48^\circ$