Question

Question 3
The angel 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.

Answer 1

Let the a parallelogram be ABCD in which $\angle ADC$ and $\angle ABC$ are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is ${60}^{\circ }$

Now BEDF is a quadrilateral, in which $\angle BED=\angle BFD={90}^{\circ }$
$\therefore \angle FBE={360}^{\circ }-(\angle FDE+\angle BED+\angle BFD)$
$={360}^{\circ }-({60}^{\circ }+{90}^{\circ }+{90}^{\circ })$
$={360}^{\circ }-{240}^{\circ }={120}^{\circ }$
Since, ABCD is a parallelogram.
$\therefore \angle ADC={120}^{\circ }$
Now, $\angle A+\angle B={180}^{\circ }$ [sum of two cointerior angles is ${180}^{\circ }]$
$\therefore \angle A={180}^{\circ }-\angle B$
$={180}^{\circ }-{120}^{\circ }[\because \angle FBE=\angle B]$
$\Rightarrow \angle A={60}^{\circ }$
Also, $\angle C=\angle A={60}^{\circ }$
Hence, angles of the parallelogram are ${60}^{\circ },{120}^{\circ },{60}^{\circ }and{120}^{\circ }$, respectively.