Question

Question 178
The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is ${45}^{\circ }$. Find the angles of the parallelogram.

Answer 1

Let ABCD be a parallelogram, where BE and BF are the perpendicular through the vertex B to the sides DC and AD, respectively.

Let $\angle A=\angle C=x,\angle B=\angle D=y$ [$\because$ opposite angles are equal in parallelogram]
Now, $\angle A+\angle B={180}^{\circ }$ [$\because$ adjacent sides of a parallelogram are supplementary]
$\Rightarrow x+\angle ABF+\angle FBE+\angle EBC={180}^{\circ }$
$\Rightarrow x+{90}^{\circ }-x+{45}^{\circ }+{90}^{\circ }-x={180}^{\circ }$
$[\because In\Delta ABF,\angle ABF={90}^{\circ }-xandin\Delta BEC,\angle EBC={90}^{\circ }-x]$
$\Rightarrow -x={180}^{\circ }-{225}^{\circ }\Rightarrow x={45}^{\circ }\therefore \angle A=\angle C={45}^{\circ }\angle B={45}^{\circ }+{45}^{\circ }+{45}^{\circ }={135}^{\circ }\Rightarrow \angle D={135}^{\circ }$
Hence the angles are ${45}^{\circ },{135}^{\circ },{45}^{\circ },{135}^{\circ }$.