The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

Name: The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.
Uploaded: Thu, 2021-01-07 15:53
Description: The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

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

Answers (1)

Given that the angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60^{\circ}$ .

In this figure $\angle ADC$ and $\angle ABC$ are obtuse angles and DE and DF are altitudes.
In quadrilateral BEDF, $\angle BED$ = $\angle BFD$ = $90^{\circ}$
$\angle FBE=360^{\circ}-(\angle FDE+\angle BED+\angle BFD)$
$=360^{\circ}-(60^{\circ}+90^{\circ}+90^{\circ})$
$=360^{\circ}-240^{\circ}$
$=120^{\circ}$
It is given that ABCD is a parallelogram
ADC = $120^{\circ}$
$\angle A+\angle B=180^{\circ}$ {Sum of interior angle $=180^{\circ}$ }
$\angle A=180^{\circ}-B$
$\angle A=180^{\circ}-120^{\circ}$
$\angle A=60^{\circ}$
$\Rightarrow \angle C=\angle A=60^{\circ}$
Hence angles of the parallelogram are $60^{\circ}, 120^{\circ}$