Chapter 4.4, Problem 28WE

To determine

Value of x and y.

Answer to Problem 28WE

Value of $x={40}^o\text{and}y={20}^o$

Explanation of Solution

Given:

In an equilateral triangle $\Delta DEF$

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

As the $\Delta DEF$ is an equilateral triangle.

So all the angles are equal to 60 degrees.

So,

  $\begin{array}{l}m\angle D=60\\ \Rightarrow x+y=60\\ \Rightarrow y=60-x\mathrm{......}\left(1\right)\end{array}$

Also

  $\begin{array}{l}m\angle E=60\\ \Rightarrow 2x-y=60\mathrm{.......}\left(2\right)\end{array}$

Substituting the value of y in (2)

  $\begin{array}{l}2x-y=60\\ 2x-\left(60-x\right)=60\\ 3x=120\\ \Rightarrow x={40}^o\end{array}$

And substituting the value of x in (1)

  $\begin{array}{l}y=60-x\\ y=60-40\\ \Rightarrow y={20}^o\end{array}$

Therefore, $x={40}^o\text{and}y={20}^o$