Question

In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If $\angle x:\angle y=2:1$ ; find angles of the parallelogram.

• A. $\angle DAB=\angle C={160}^{\circ }$ and $\angle B=\angle D={60}^{\circ }$
• B. $\angle DAB=\angle C={120}^{\circ }$ and $\angle B=\angle D={60}^{\circ }$
• C. $\angle DAB=\angle C={230}^{\circ }$ and $\angle B=\angle D={60}^{\circ }$
• D. $\angle DAB=\angle C={100}^{\circ }$ and $\angle B=\angle D={60}^{\circ }$

Answer 1

The correct option is B $\angle DAB=\angle C={120}^{\circ }$ and $\angle B=\angle D={60}^{\circ }$

Given, ABCD is a parallelogram and $AQ\perp CDandAP\perp BC$
$\therefore \angle AQC=\angle APC=90$
AQCP is forming a quadrilateral.
As we know that sum of angles of quadrilateral is 360,
$\therefore \angle QAP+\angle AQC+\angle QCP+\angle APC=360$
$or,\angle QAP+90+\angle QCP+90=360(\because \angle AQC=\angle APC=90)or,\angle QAP+\angle QCP=360-180or,\angle QAP+\angle QCP=180\left(1\right)$
Given $\angle QCP=x,\angle QAP=yx:y=2:1$
Let angle be m.
$\therefore x=2mandy=m$
from (1),
$\angle QAP+\angle QCP=180or,m+2m=180or,3m=180or,m=60$
$\angle QCP=2mor,\angle C=2\times 60=120\left(2\right)$
In a parallelogram ABCD, Sum of any two consecutive angles is supplementary,
$\therefore \angle C+\angle D=180or,\angle 120+\angle D=180or,\angle D=180-120or,\angle D=60\left(3\right)$
In a paralleogram ABCD, opposite angles are equal,
$\therefore \angle C=\angle A=120\left(from2\right)\angle D=\angle B=60\left(from3\right)$