The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.
Draw a parallelogram $\mathrm{ABCD}$.
Drop a perpendicular from $\mathrm{B}$ to the side $\mathrm{AD}$, at the point $\mathrm{E} .$
Drop perpendicular from $\mathrm{B}$ to the side $\mathrm{CD}$, at the point $\mathrm{F}$.
In the quadrilateral $\mathrm{BEDF}:$
$\angle \mathrm{EBF}=60^{\circ}, \angle \mathrm{BED}=90^{\circ}$
$\angle \mathrm{BFD}=90^{\circ}$
$\angle \mathrm{EDF}=360^{\circ}-\left(60^{\circ}+90^{\circ}+90^{\circ}\right)=120^{\circ}$
In a parallelogram, opposite angles are congruent and adjacent angles are supplementary.
In the parallelogram $\mathrm{ABCD}$ :
$\angle \mathrm{B}=\angle \mathrm{D}=120^{\circ}$
$\angle \mathrm{A}=\angle \mathrm{C}=180^{\circ}-120^{\circ}=60^{\circ}$