Question

In a parallelogram ABCD, AP and AQ are perpendicular 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 degree)
a) ${{55}^{\circ }}$
b) ${{60}^{\circ }}$
c) ${{70}^{\circ }}$
d) ${{48}^{\circ }}$

Answer 1

Hint: Consider the quadrilateral in fig. QAPC, by using angle sum find the value of x and y in the figure. In a parallelogram opposite angles are equal. Thus find the smallest angle.

Complete step-by-step answer:
Given to us a parallelogram ABCD. It is said that AD and AQ are perpendicular from the vertex of angle A. Hence from the figure we can say that AQ is perpendicular to CD and AP is perpendicular to BC.
$\begin{align}
  & AQ\bot CD,AP\bot BC \\
 & \Rightarrow \angle AQC=\angle APC={{90}^{\circ }} \\
\end{align}$
From the figure AQCP is in the form of quadrilateral, we know that sum of the angles in a quadrilateral is ${{360}^{\circ }}$ . Hence, from the figure taking the quadrilateral AQCP, we can say that
$\angle QAP+\angle AQC+\angle QCP+\angle APC={{360}^{\circ }}$
Now we know that,
$\angle AQC=\angle APC={{90}^{\circ }}$
Let us substitute this in the above expression
$\begin{align}
  & \angle QAP+{{90}^{\circ }}+\angle QCP+{{90}^{\circ }}={{360}^{\circ }} \\
 & \therefore \angle QAP+\angle QCP={{360}^{\circ }}-{{180}^{\circ }} \\
 & ={{180}^{\circ }} \\
 & \therefore \angle QAP+\angle QCP+={{180}^{\circ }}.........................\left( i \right) \\
\end{align}$
We have been given,
$\angle QAP=\angle y,\angle QCP=\angle x$
It is given that x : y = 2 : 1. Let us consider that angle to be m. thus x = 2m and y = m.
\[\begin{align}
  & \angle QAP+\angle QCP={{180}^{\circ }} \\
 & m+2m={{180}^{\circ }} \\
 & 3m={{180}^{\circ }}\Rightarrow m=\dfrac{{{180}^{*}}}{3}={{60}^{\circ }} \\
 & \Rightarrow \angle QAP=y={{60}^{\circ }} \\
 & \angle QCP=2m=2\times {{60}^{\circ }}={{120}^{\circ }} \\
 & \therefore \angle A={{60}^{\circ }},\angle C={{120}^{\circ }} \\
\end{align}\]

Now in a parallelogram ABCD, opposite angles are equal. Thus,
$\begin{align}
  & \angle C=\angle A={{120}^{\circ }}, \\
 & \angle D=\angle B={{60}^{\circ }} \\
\end{align}$
Thus the smallest angle that the parallelogram has is $\angle D,\angle B={{60}^{\circ }}$ . The smallest angle of the parallelogram is ${{60}^{\circ }}$ .
So, option (b) is correct.

Note:It is important that we have to identify QAPC as a quadrilateral, we should know the basic properties of parallelograms to figure out that their opposite angles are equal.