Chapter 4.4, Problem 29WE

To determine

Value of x and y.

Answer to Problem 29WE

Value of $x={30}^o\text{and}y={10}^o$

Explanation of Solution

Given:

In $\Delta JKL,\overline{JK}\cong \overline{KL}$

  $m\angle D=x+y\text{and }m\angle E=2x-y$

According to the given condition the triangle is

  

Here. $\overline{JK}\cong \overline{KL}$

So, $\Delta JKL$ is an isosceles triangle.

So, from the property of an isosceles triangle

  $\begin{array}{l}m\angle J=m\angle L\\ \Rightarrow 2x-y=x+2y\\ \Rightarrow x=3y\mathrm{.....}\left(1\right)\end{array}$

Also. The sum of angles of a triangle is 180 degrees.

So,

  $\begin{array}{l}m\angle J+m\angle K+m\angle L=180\\ \Rightarrow \left(2x-y\right)+\left(2x+2y\right)+\left(x+2y\right)=180\\ \Rightarrow 5x+3y=180\mathrm{.....}\left(2\right)\end{array}$

Substituting the value of x in (2)

  $\begin{array}{l}5x+3y=180\\ \Rightarrow 5\left(3y\right)+3y=180\\ \Rightarrow 18y=180\\ \Rightarrow y={10}^o\end{array}$

And substituting the value of y in (1)

  $\begin{array}{l}x=3y\\ \Rightarrow x=3\times 10\\ \Rightarrow x={30}^o\end{array}$

Therefore, $x={30}^o\text{and}y={10}^o$