Chapter 5.5, Problem 19PSB

To determine

To calculate: the angle at $K$ vertex in the given parallelogram $KMOP$

Answer to Problem 19PSB

The angle at vertex $\left(\angle K\right)=28°$

Explanation of Solution

Given information:

  $▱KMOP$ is a parallelogram.

  $\begin{array}{l}\angle M=\left(x+3y\right)°\\ \angle O=\left(x-4\right)°\\ \angle P=\left(4y-8\right)°\end{array}$

Formula used:

• Opposite angles in a parallelogram are equal.
• Adjacent angles in a parallelogram are supplementary.

Calculation:

  $\begin{array}{l}\angle M=\angle P\\ \left[\because \text{Opposite angles are equal}\right]\\ x+3y=4y-8\\ x+3y-4y=-8\\ x-y=-8\dots \left(i\right)\end{array}$

  

  $\begin{array}{l}\angle O+\angle P=180°\\ \left[\because \text{Adjacent angles are supplementary}\right]\\ \left(x-4\right)+\left(4y-8\right)=180\\ x-4y-12=180\\ x-4y=192\dots \left(ii\right)\end{array}$

  

Solving $\left(i\right)\text{ and }\left(ii\right)$ , we get

  $\begin{array}{l}x-y=-8\\ x+4y=192\\ \_\_\_\_\_\_\_\_\_\_\_\\ -5y=-200\\ y=\frac{200}{5}\\ y=40\end{array}$

On substituting $y=40$ in $\left(i\right)$ ,

  $\begin{array}{l}\Rightarrow x-40=-8\\ \Rightarrow x=40-8\\ \Rightarrow x=32\end{array}$

Now,

  $\begin{array}{l}\angle K=x-4\\ \Rightarrow \angle K=32-4\\ \Rightarrow \angle K=28\end{array}$