Chapter 12, Problem 35CR

To determine

To find: The measure of an arc $CD.$

Answer to Problem 35CR

The measure of arc $CD\text{is}{32}^{\text{o}}.$

Explanation of Solution

Given Information:

  $m\stackrel{⌢}{AB}:m\stackrel{⌢}{CD}=5:2$

  $\angle P={24}^{\text{o}}$

Formula used:

If $PDA\text{and}PCB$ are two secants intersecting outsides the circle, then the measure of the angle between them is equal to one-half the difference of the measure of the intercepted arcs.

i.e. $\angle P=\frac{1}{2}\left(m\stackrel{⌢}{AB}-m\stackrel{⌢}{CD}\right)$

Calculation:

As $m\stackrel{⌢}{AB}:m\stackrel{⌢}{CD}=5:2$

Let $m\stackrel{⌢}{AB}=5x\text{and}m\stackrel{⌢}{CD}=2x$

Now, using the property of two secants intersecting outside the circle, we have

  $\angle P=\frac{1}{2}\left(m\stackrel{⌢}{AB}-m\stackrel{⌢}{CD}\right)$

  $\begin{array}{l}\therefore {24}^{\text{o}}=\frac{1}{2}\left(5x-2x\right)\\ \therefore {48}^{\text{o}}=3x\\ \therefore x={16}^{\text{o}}\\ \therefore m\stackrel{⌢}{CD}=2x={32}^{\text{o}}\end{array}$

Hence, the measure of arc $CD\text{is}{32}^{\text{o}}.$