Chapter 5.7, Problem 3PSA

To determine

To calculate:Thevalue ofxforRECT to be rectangle.

Answer to Problem 3PSA

The value of xis${21}^{\circ }$ .

Explanation of Solution

Given information:

In RECT,

  $\begin{array}{l}\angle ETC={42}^{\circ },\\ \angle RTE={\left(2x+6\right)}^{\circ }.\end{array}$

Calculation:

  

In RECT,

  $\begin{array}{l}\angle ETC={42}^{\circ },\\ \angle RTE={\left(2x+6\right)}^{\circ }.\end{array}$

In rectangle all angles are right angles.

The measure of angle T is ${90}^{\circ }$

  $\begin{array}{l}\angle ETC+\angle RTE={90}^{\circ }\\ {42}^{\circ }+{\left(2x+6\right)}^{\circ }={90}^{\circ }\\ {42}^{\circ }+2x+6^{\circ }={90}^{\circ }\\ 2x+{48}^{\circ }={90}^{\circ }\\ 2x={90}^{\circ }-{48}^{\circ }\\ 2x={42}^{\circ }\\ x=\frac{{42}^{\circ }}{2}={21}^{\circ }\end{array}$