Chapter 4.4, Problem 33WE

To determine

To calculate: The measure of angles $\angle EAF\text{ , }\angle \text{AFD and }\angle \text{DAF}$ .

Answer to Problem 33WE

  $\begin{array}{l}m\angle EAF=9^{\circ }\\ m\angle AFD={54}^{\circ }\\ m\angle DAF={45}^{\circ }\end{array}$

Explanation of Solution

Given information: 

In the figure shown,

$\begin{array}{l}AB=BC=CD=DE=AE=EF=FG=DG\\ \angle BAE=\angle ABC=\angle BCD=\angle CDE=\angle DEA={108}^{\circ }\\ \angle DEF=\angle EFG=\angle FGD=\angle GDE={90}^{\circ }\end{array}$

Calculation:

In $\Delta AEF$ , $AE=EF$ .

Thus, $\Delta AEF$ is an isosceles triangle.

In an isosceles triangle, angles opposite to equal sides are equal.

So, $m\angle EAF=m\angle EFA$ .

  $\begin{array}{l}m\angle AEF+m\angle AED+m\angle FED={360}^{\circ }\\ m\angle AEF+{108}^{\circ }+{90}^{\circ }={360}^{\circ }\\ m\angle AEF={360}^{\circ }-{198}^{\circ }\\ m\angle AEF={162}^{\circ }\end{array}$

Now,

  $\begin{array}{l}m\angle AEF+m\angle EAF+m\angle EFA={180}^{\circ }\\ {162}^{\circ }+m\angle EAF+m\angle EAF={180}^{\circ }\\ 2m\angle EAF={18}^{\circ }\\ m\angle EAF=9^{\circ }\end{array}$

So,

  $m\angle EFA=m\angle EAF=9^{\circ }$ .

Now,

  $\begin{array}{l}m\angle EFA+m\angle EFD=m\angle AFD\\ 9^{\circ }+{45}^{\circ }=m\angle AFD\\ m\angle AFD={54}^{\circ }\end{array}$

In $\Delta AED$ ,

  $\begin{array}{l}AE=ED\\ m\angle DAE=m\angle EDA\\ m\angle DAE+m\angle EDA+m\angle AED={180}^{\circ }\\ m\angle DAE+m\angle DAE+{108}^{\circ }={180}^{\circ }\\ 2m\angle DAE={72}^{\circ }\\ m\angle DAE={36}^{\circ }\\ \\ m\angle DAE+m\angle EAF=m\angle DAF\\ m\angle DAF={36}^{\circ }+9^{\circ }\\ m\angle DAF={45}^{\circ }\end{array}$