Chapter 53, Problem 20A

Solve the following exercises based on Principles 11-14, although an exercise may require the application of two or more of any of the principles. Round the answers to 3 decimal places where necessary unless otherwise stated.
a. If $AB=5.378"$ and $AC=3.782"$ , find (1) DB (2) $\stackrel{⌢}{ACB.}$

b If $DB=3.017"$ and $\stackrel{⌢}{ACB}=7.308"$ , find (1) AB (2) $\stackrel{⌢}{CB.}$

To determine

(a)

The value of DB and arc $\stackrel{⌢}{ACB}$.

Answer to Problem 20A

The value of DB is $DB=\mathrm{2.689.}$

Explanation of Solution

Given information:

AB = 5.378"

  $\stackrel{⌢}{AC}=3.782"$

  

If the radius of the circle is perpendicular to the chord, it always divides the chord in two equal parts.

In the figure given above, the radius is perpendicular to the chord; hence, the parts AD will be equal to DB. Each portion AD and DB is half the value of part AB.

  $AD=DB=\frac{AB}{2}$

  $\begin{array}{l}DB=\frac{AB}{2}=\frac{5.378}{2}=2.689\\ DB=2.689\end{array}$

The value of DB is $DB=2.689$

If the radius of the circle is perpendicular to the chord, it always divides the chord and the arc cut off by the chord in two equal parts.

In the figure given above, the radius is perpendicular to the chord; hence, the parts AD will be equal to DB. Each portion AD and DB is half the value of part AB. Also, the arc cut off by the chord is divided in two equal parts.

  $\stackrel{⌢}{AC}=\stackrel{⌢}{CB}=\frac{\stackrel{⌢}{ACB}}{2}$

  $\begin{array}{l}\stackrel{⌢}{AC}=\stackrel{⌢}{CB}=\frac{\stackrel{⌢}{ACB}}{2}\\ \stackrel{⌢}{AC}=3.782"\\ \stackrel{⌢}{ACB}=2\times 3.782"=7.564"\\ \stackrel{⌢}{ACB}=7.564"\end{array}$

The value of DB is $DB=2.689$ and the value of arc is $\stackrel{⌢}{ACB}=7.564".$

Conclusion:

Thus, the value of DB is $DB=2.689$ and the value of arc is $\stackrel{⌢}{ACB}=7.564".$

To determine

(b)

The value of AB and arc $\stackrel{⌢}{CB}$.

Answer to Problem 20A

The value of AB is $AB=6.034"$ and the value of arc is $\stackrel{⌢}{CB}=3.654".$

Explanation of Solution

Given information:

AB = 3.017"

  $\stackrel{⌢}{ACB}=7.308"$

  

If the radius of the circle is perpendicular to the chord, it always divides the chord in two equal parts.

In the figure given above, the radius is perpendicular to the chord; hence, the parts AD will be equal to DB. Each portion AD and DB is half the value of part AB.

  $AD=DB=\frac{AB}{2}$

  $\begin{array}{l}AB=2\times DB\\ AB=2\times 3.017"\\ AB=6.034"\end{array}$

The value of AB is $AB=6.034"$

If the radius of the circle is perpendicular to the chord, it always divides the chord and the arc cut off by the chord in two equal parts.

In the figure given above, the radius is perpendicular to the chord; hence, the parts AD will be equal to DB. Each portion AD and DB is half the value of part AB. Also, the arc cut off by the chord is divided in two equal parts.

  $\stackrel{⌢}{AC}=\stackrel{⌢}{CB}=\frac{\stackrel{⌢}{ACB}}{2}$

  $\begin{array}{l}\stackrel{⌢}{AC}=\stackrel{⌢}{CB}=\frac{\stackrel{⌢}{ACB}}{2}\\ \stackrel{⌢}{ACB}=7.308"\\ \stackrel{⌢}{CB}=\frac{7.308"}{2}\\ \stackrel{⌢}{CB}=3.654"\end{array}$

The value of AB is $AB=6.034"$ and the value of arc is $\stackrel{⌢}{CB}=3.654"$

Conclusion:

Thus, the value of AB is $AB=6.034"$ and the value of arc is $\stackrel{⌢}{CB}=3.654".$