Chapter 11.4, Problem 17WE

(a)

To determine

To Explain: Why $m\angle AOX=18$

Answer to Problem 17WE

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

  $\begin{array}{l}m\angle AOB=\frac{360}{n}\\ m\angle AOX=\frac{1}{2}m\angle AOB\\ m\angle AOX=\frac{1}{2}\left(\frac{360}{10}\right)\\ m\angle AOX=18\end{array}$

The measure of inscribed angle is half of central angle.

(b)

To determine

To find:

  $\begin{array}{l}\sin {18}^{°}=\frac{AX}{1}=AX=\mathrm{...}\\ \cos {18}^{°}=\frac{OX}{1}=OX=\mathrm{...}\end{array}$

Answer to Problem 17WE

  $\begin{array}{l}OX=0.9510\\ AX=0.3090\end{array}$

Explanation of Solution

Given:

  

  $\begin{array}{l}\sin {18}^{°}=\frac{AX}{1}\\ 0.3090=\frac{AX}{1}\\ AX=0.3090\\ \cos {18}^{°}=\frac{OX}{1}\\ 0.9510=\frac{OX}{1}\\ OX=0.9510\end{array}$

Hence,

  $\begin{array}{l}OX=0.9510\\ AX=0.3090\end{array}$

(c)

To determine

To find: Perimeter of decagon.

Answer to Problem 17WE

  $P=6.18cm$

Explanation of Solution

Given:

  

  $\begin{array}{l}AX=\frac{AB}{2}=\frac{s}{2}\\ s=2AX\\ P=10s\\ P=10(2AX)\\ P=20(0.3090)\\ P=6.18cm\end{array}$

(d)

To determine

To find: Area of $AOB$ .

Answer to Problem 17WE

  $A=0.293$

Explanation of Solution

Given:

  

  $\begin{array}{l}A=\frac{1}{2}ap\\ A=\frac{1}{2}(2AX)(OX)\\ A=(AX)(OX)\\ A=(0.3090)(0.9510)\\ A=0.293\end{array}$

(e)

To determine

To Find: Area of decagon.

Answer to Problem 17WE

  $A=2.93$

Explanation of Solution

Given:

  

  $\begin{array}{l}A=10(ar(\Delta AOB)\\ A_{AOB}=0.2939\\ thus\\ A=10(0.2939)\\ A=2.93\end{array}$