Question

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

• A. ${55}^o$
• B. ${60}^o$
• C. ${70}^o$
• D. ${48}^o$

Answer 1

The correct option is B ${60}^o$

Given is a parallelogram ABCD. AP & AQ are perpendiculars drawn from vertex A.

So $\angle AQD=\angle AQC=\angle APC=\angle PAB={90}^o$

Also given $\angle C=x$ & $\angle QAP=y$ and $x:y=2:1$

Now in quadrilateral QCPA,

$\angle QAP+\angle APC+\angle PCQ+\angle CQA={360}^o$

$\Rightarrow \angle QAP+\angle PCQ+{90}^o+{90}^o={360}^o$

$\Rightarrow \angle QAP+\angle PCQ={180}^o$

$\Rightarrow y+x={180}^o$

Since $x:y=2:1$ so $x=\frac{2}{3}\times {180}^o={120}^o$

$\Rightarrow \angle C={120}^o$

So $\angle A=\angle C={120}^o$

Now $\angle C+\angle B={180}^o$ [ adjacent angles in parallelogram are supplementary ]

$\Rightarrow \angle B={180}^o-{120}^o={60}^o$

So $\angle B=\angle D={60}^o$