The angle between two altitudes of a parallelogram drawn from the vertex of an obtuse angle is $60^{\circ}$. Find the angles of the parallelogram.

(N/A) Let the parallelogram be $ABCD$. Let $DP \perp AB$ and $DQ \perp BC$ be the two altitudes from the obtuse vertex $D$.
In quadrilateral $DPBQ$,the sum of the interior angles is $360^{\circ}$.
$\angle PDQ + \angle DPB + \angle B + \angle DQB = 360^{\circ}$
Given $\angle PDQ = 60^{\circ}$,$\angle DPB = 90^{\circ}$,and $\angle DQB = 90^{\circ}$.
$60^{\circ} + 90^{\circ} + \angle B + 90^{\circ} = 360^{\circ}$
$\angle B + 240^{\circ} = 360^{\circ}$
$\angle B = 360^{\circ} - 240^{\circ} = 120^{\circ}$.
Since opposite angles of a parallelogram are equal,$\angle D = \angle B = 120^{\circ}$.
Since consecutive interior angles are supplementary,$\angle A + \angle B = 180^{\circ}$.
$\angle A + 120^{\circ} = 180^{\circ} \Rightarrow \angle A = 60^{\circ}$.
Since opposite angles are equal,$\angle C = \angle A = 60^{\circ}$.
Thus,the angles of the parallelogram are $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.