Chapter 12, Problem 28CR

a.

To determine

To find: The measure of arc $AD$ if side CBis $9$ .

Answer to Problem 28CR

The measure of arc $AD$ is $120°$ .

Explanation of Solution

Given information:

In a circle with center O,

Side $CB=9,$

Side $\angle C={30}^{\circ }.$

  $\overline{BC}\cong \overline{BD}$ .

  $\overleftrightarrow{CD}$ is tangent to circle with center O.

Calculation:

  

A sum of angles in triangle is ${180}^{\circ }.$

  $\begin{array}{l}m\angle D+m\angle C+m\angle DBC={180}^{\circ }\\ {30}^{\circ }+{30}^{\circ }+m\angle DBC={180}^{\circ }\\ m\angle DBC={180}^{\circ }-{60}^{\circ }={120}^{\circ }\\ m\angle ABD={180}^{\circ }-m\angle DBC\\ m\angle ABD={180}^{\circ }-{120}^{\circ }={60}^{\circ }\\ m\angle ABD=\frac{1}{2}\left(\stackrel{⌢}{AD}\right)\\ {60}^{\circ }=\frac{1}{2}\left(\stackrel{⌢}{AD}\right)\\ m\left(\stackrel{⌢}{AD}\right)={120}^{\circ }\end{array}$

b.

To determine

To calculate: The length CDwhich is tangent to circle with center O.

Answer to Problem 28CR

The length CDis$15.58$ .

Explanation of Solution

Given information:

In a circle with center O,

Side $CB=9,$

Side $\angle C={30}^{\circ }.$

  $\overline{BC}\cong \overline{BD}$ .

  $\overleftrightarrow{CD}$ is tangent to circle with center O.

Calculation:

  

Altitude $\overline{BX}$ to the base of isosceles triangle DBC forms two ${30}^{\circ }{60}^{\circ }{90}^{\circ }$ triangles.

  $\begin{array}{l}BX=\frac{1}{2}\left(9\right)\\ BX=4.5\\ CX=BX\sqrt{3}\\ CX=4.5\sqrt{3}=7.79\\ CD=2CX\\ CD=\left(2\times 7.79\right)=15.58\end{array}$

c.

To determine

To calculate: The radius of circle with center O.

Answer to Problem 28CR

The area of circleis $9$ .

Explanation of Solution

Given information:

In a circle with center O,

Side $CB=9,$

Side $\angle C={30}^{\circ }.$

  $\overline{BC}\cong \overline{BD}$ .

  $\overleftrightarrow{CD}$ is tangent to circle with center O.

Calculation:

  

We use the below theorem:

Tangent-Secant Power Theorem: If a tangent and secant are drawn from an external point to a circle, then the square of the length of the tangent is equal to the product of the length of the secant’s external part and the length of the entire secant.

By applyingTangent-Secant Power Theorem, we get

  $\begin{array}{l}{\left(CD\right)}^2=\left(BC\right)\left(AC\right)\\ {\left(15.58\right)}^2=\left(9\right)\left(AC\right)\\ 225=\left(9\right)\left(9+y\right)\\ 243=81+9y\\ 162=9y\\ y=\frac{162}{9}\\ y=18\\ r=\frac{y}{2}=\frac{18}{2}=9\end{array}$