The angle between two altitudes


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


Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Now, $B E D F$ is a quadrilateral, in which $\angle B E D=\angle B F D=90^{\circ}$

$\therefore$  $\angle F B E=360^{\circ}-(\angle F D E+\angle B E D+\angle B F D)$





Since, $A B C D$ is a parallelogram.

$\therefore$ $\angle A D C=120^{\circ}$

Now, $\angle A+\angle B=180^{\circ}$ [sum of two cointerior angles is $180^{\circ}$ ]

$5=180^{\circ}-120^{\circ}$ $[\because \angle F B E=\angle B]$

$\Rightarrow$ $\angle A=60^{\circ}$ 

Also,  $\angle C=\angle A=60^{\circ}$

Hence, angles of the parallelogram are $60^{\circ}, 120^{\circ}, 60^{\circ}$ and $120^{\circ}$, respectively.




Leave a comment