Question

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

Answer 1

In parallelogram ABCD,

$\angle isanobtusedangleAE\perp BCandAF\perp DC\angle EAF={60}^{\circ }$
In quadrilateral AECF,
$\angle EAF+\angle F+\angle D+\angle E={360}^{\circ }$
(Sum of angles of a quadrilateral)
$\Rightarrow {60}^{\circ }+{90}^{\circ }+\angle C+{90}^{\circ }={360}^{\circ }\Rightarrow {240}^{\circ }+\angle C={360}^{\circ }\Rightarrow \angle C={360}^{\circ }-{240}^{\circ }={120}^{\circ }\therefore \angle C={120}^{\circ }$
But $\angle A=\angle C$ (Opposite angles)
$\therefore \angle A={120}^{\circ }$
But $\angle A+\angle B={180}^{\circ }$
(Sum of adjacent angles)
$\Rightarrow {120}^{\circ }+\angle B={180}^{\circ }\angle B={180}^{\circ }-{120}^{\circ }={60}^{\circ }\therefore \angle B={60}^{\circ }$
But $\angle D=\angle B$ (Opposite angles)
$\therefore \angle D={60}^{\circ }$
Hence $\angle A={120}^{\circ },\angle B={60}^{\circ },\angle C={120}^{\circ }and\angle D={60}^{\circ }$