Question

If an angle of a parallelogram is two – third of its adjacent angle, find the angles of a parallelogram.

Answer 1

Hint: We will first start by using the fact that the sum of adjacent angles of a parallelogram is $180{}^\circ $. Then we will let the one angle of parallelogram as x and according to the question we have the adjacent angle as $\dfrac{2}{3}x$. Then we will use the fact to find the value of x and similarly other angles.

Complete step-by-step answer:
Now, we have been given that one angle of a parallelogram is $\dfrac{2}{3}$ of its adjacent. Therefore, we let the one angle of parallelogram as x. Therefore, the adjacent angle is $\dfrac{2}{3}x$.

Now, we know that the sum of the adjacent angles of a parallelogram is $180{}^\circ $. Therefore, we have,
$\begin{align}
  & x+\dfrac{2}{3}x=180{}^\circ \\
 & \dfrac{5x}{3}=180{}^\circ \\
 & x=\dfrac{3}{5}\times 180{}^\circ \\
 & x=3\times 36{}^\circ \\
 & x=108{}^\circ \\
 & \dfrac{2}{3}x=72{}^\circ \\
\end{align}$
Now, we know that the opposite angles of a parallelogram are equal. Therefore, we have,
\[\begin{align}
  & \angle ABC=\dfrac{2}{3}x=72{}^\circ \\
 & \angle BCD=x=108{}^\circ \\
\end{align}\]
The angles of the parallelogram are \[108{}^\circ ,72{}^\circ ,108{}^\circ ,72{}^\circ \] respectively.

Note: We have used a property of parallelogram that the opposite angles of a parallelogram are equal. Therefore, we have,
\[\begin{align}
  & \angle ABC=\angle ADC=72{}^\circ \\
 & \angle BAD=\angle BCD=108{}^\circ \\
\end{align}\]
Using this property has helped us to find all the angles of the parallelogram easily.