Chapter 11.4, Problem 9CE

(a)

To determine

To Find: $m\angle AOB$

Answer to Problem 9CE

  $m\angle AOB={72}^{°}$

Explanation of Solution

Given:

  

Using central angle property

  $\begin{array}{l}m\angle AOB=\frac{360}{n}\\ n=5\\ m\angle AOB=\frac{360}{5}\\ m\angle AOB={72}^{°}\end{array}$

Here $n$ is the number of triangles formed.

(b)

To determine

To Explain: Why $m\angle AOX={36}^{°}$

Answer to Problem 9CE

The measure of inscribed angle is half of central angle.

Explanation of Solution

Given:

  

Using central angle

  $m\angle AOB=\frac{360}{n}$

Here $n$ is the number of triangles formed.

Also

  $\begin{array}{l}m\angle AOB=\frac{360}{n}\\ m\angle AOX=\frac{1}{2}m\angle AOB\\ m\angle AOX=\frac{1}{2}\left(\frac{360}{5}\right)\\ m\angle AOX={36}^{°}\end{array}$

The measure of inscribed angle is half of central angle.

(c)

To determine

To find: $\cos {36}^{°}=\frac{a}{?}tothenearesttentha\approx ?$

Answer to Problem 9CE

  $a=8.1$

Explanation of Solution

Given:

  

  $\begin{array}{l}In\Delta AOX\\ \cos {36}^{°}=\frac{a}{10}\\ a=10\cos {36}^{°}\\ a=10(.8090)\\ a=8.1\end{array}$

(d)

To determine

To find: $\sin {36}^{°}=\frac{\frac{1}{2}s}{?}tothenearesttenths\approx ?$

Answer to Problem 9CE

  $s=11.8$

Explanation of Solution

Given:

  

  $\begin{array}{l}In\Delta AOX\\ \sin {36}^{°}=\frac{\frac{1}{2}s}{10}\\ \frac{s}{2}=10\sin {36}^{°}\\ s=20(0.5878)\\ s=11.8\end{array}$

(e)

To determine

To Find: The perimeter and Area of decagon.

Answer to Problem 9CE

  $P=59,A=238.95$

Explanation of Solution

Given:

  

  $\begin{array}{l}P=ns\\ P=5(11.8)\\ P=59\\ A=\frac{1}{2}ap\\ A=\frac{1}{2}(8.1)(59)\\ A=238.95\end{array}$