Question

question 39 of 44 what is the measure of minor arc ( widehat{ac} ), given that ( overline{ab} ) and ( overline{bc} ) are tangent to ( odot o )? a. 220° b. 200° c. 160° d. 180°

Explanation:

Step1: Recall tangent - circle angle formula

The measure of an angle formed by two tangents drawn from an external point to a circle is equal to half the difference of the measures of the intercepted arcs. Let the measure of the minor arc \(\widehat{AC}\) be \(x\) and the measure of the major arc \(\widehat{ADC}\) be \(y\). We know that the sum of the measures of a major arc and a minor arc that make up a circle is \(360^{\circ}\), so \(x + y=360^{\circ}\). Also, the measure of \(\angle B\) (the angle formed by the two tangents) is given as \(20^{\circ}\), and by the tangent - angle formula, \(\angle B=\frac{1}{2}(y - x)\).

Step2: Substitute known values

We are given that the major arc \(\widehat{ADC}\) has a measure of \(200^{\circ}\)? Wait, no, wait. Wait, the diagram shows arc \(ADC\) as \(200^{\circ}\)? Wait, no, let's re - check. Wait, the angle at \(B\) is \(20^{\circ}\). Let's use the formula \(\angle B=\frac{1}{2}(\text{measure of major arc}-\text{measure of minor arc})\). Let the minor arc \(\widehat{AC}=x\), major arc \(\widehat{ADC}=360 - x\). Then \(20^{\circ}=\frac{1}{2}((360 - x)-x)\).

Step3: Solve the equation

First, multiply both sides of the equation \(20=\frac{1}{2}(360 - 2x)\) by 2: \(40 = 360-2x\). Then, add \(2x\) to both sides: \(2x + 40=360\). Subtract 40 from both sides: \(2x=360 - 40=320\). Then divide both sides by 2: \(x = 160^{\circ}\). Wait, but wait, maybe I misread the arc. Wait, the diagram has arc \(ADC\) as \(200^{\circ}\)? Wait, no, the problem says "given that \(\overline{AB}\) and \(\overline{BC}\) are tangent to \(\odot O\)". Wait, maybe the major arc is \(200^{\circ}\)? Wait, no, let's re - derive.

The formula for the angle between two tangents: \(\angle B=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Let minor arc \(AC = x\), major arc \(AC\) (the other arc) \(=360 - x\). We know that \(\angle B = 20^{\circ}\), so \(20=\frac{1}{2}((360 - x)-x)\) → \(20=\frac{1}{2}(360 - 2x)\) → \(40 = 360-2x\) → \(2x=320\) → \(x = 160\). But wait, the diagram shows arc \(ADC = 200^{\circ}\). Wait, maybe the major arc is \(200^{\circ}\)? No, that can't be. Wait, no, if arc \(ADC\) is \(200^{\circ}\), then the minor arc \(AC=360 - 200 = 160^{\circ}\), and the angle at \(B\) should be \(\frac{1}{2}(200 - 160)=\frac{1}{2}(40) = 20^{\circ}\), which matches the given angle. So the measure of minor arc \(AC\) is \(160^{\circ}\).

Answer:

C. 160°