Chapter 4.6, Problem 42PPS

To determine

The value of the $m\angle JLK$

Answer to Problem 42PPS

  $m\angle JLK=20$

Explanation of Solution

Given:

The given figure is:

  

Calculation:

Angle LPM and angle LPQ are supplementary. Therefore,

  $\begin{array}{l}m\angle LPM+m\angle LPQ=180\\ 3x-5+2x+10=180\\ 5x-45=180\\ 5x=225\\ x=\frac{225}{5}\\ x=45\end{array}$

Angle LPM will be:

  $\begin{array}{l}m\angle LPM=3x-55\\ m\angle LPM=3\left(45\right)-55\\ m\angle LPM=80\end{array}$

As triangle is an isosceles, therefore,

  $\begin{array}{c}m\angle LMP=m\angle LPM\\ m\angle LMP=80\end{array}$

The value of angle MLP will be:

  $\begin{array}{c}m\angle MLP=180-m\angle LPM-m\angle LMP\\ m\angle MLP=180-80-80\\ m\angle MLP=20\end{array}$

The value of the $m\angle JLK$ will be:

  $m\angle JLK=m\angle MLP=20$