Question

secants, tangents, and angles
finding arc measures
40°
20°
satellite
o
earth
80°
160°
320°

Explanation:

Step1: Recall the tangent - secant angle formula

The measure of an angle formed by two tangents (or a tangent and a secant) drawn from an external point to a circle is half the difference of the measures of the intercepted arcs. For a tangent - tangent case (here, the two lines from the satellite to the circle are tangents, so the intercepted arcs are the major arc and the minor arc \(x\), and the sum of the major arc and minor arc is \(360^{\circ}\)). The formula is \(\theta=\frac{1}{2}( \text{major arc}-\text{minor arc})\), where \(\theta = 20^{\circ}\) (the angle at the satellite). Let the minor arc be \(x\) and the major arc be \(360 - x\).

Step2: Substitute into the formula

Substitute \(\theta = 20^{\circ}\) into the formula: \(20=\frac{1}{2}((360 - x)-x)\)

Step3: Solve the equation

First, multiply both sides by 2: \(40=(360 - x)-x\)
Simplify the right - hand side: \(40 = 360-2x\)
Subtract 360 from both sides: \(40 - 360=-2x\)
\(- 320=-2x\)
Divide both sides by \(- 2\): \(x = 160^{\circ}\)? Wait, no, wait. Wait, actually, if the angle is formed by two tangents, the formula is \(\theta=\frac{1}{2}(\text{major arc}-\text{minor arc})\), but also, the angle between two tangents and the minor arc: the measure of the angle between two tangents is half the difference of the major arc and the minor arc. But another way: the angle between two tangents is equal to \(180^{\circ}-\) the measure of the minor arc? No, no, let's re - derive.

Wait, the correct formula for the angle formed by two tangents from an external point: the measure of the angle is half the difference of the measures of the intercepted arcs. The intercepted arcs are the major arc and the minor arc. So \(\angle\text{satellite}=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Let the minor arc be \(x\), major arc be \(360 - x\). So \(20=\frac{1}{2}((360 - x)-x)\)

\(20=\frac{1}{2}(360 - 2x)\)

Multiply both sides by 2: \(40 = 360-2x\)

\(2x=360 - 40\)

\(2x = 320\)

\(x = 160^{\circ}\)? Wait, no, that can't be. Wait, maybe I mixed up. Wait, the angle between two tangents: the measure of the angle is equal to \(180^{\circ}-\) the measure of the central angle of the minor arc? No, no. Wait, let's think again. The sum of the angle at the external point and the measure of the minor arc's central angle? No, the correct formula: if two tangents are drawn from an external point, then the angle between the tangents is supplementary to the measure of the minor arc? No, no, the formula is \(\theta=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Let's take an example: if the minor arc is \(140^{\circ}\), major arc is \(220^{\circ}\), then the angle is \(\frac{1}{2}(220 - 140)=40^{\circ}\). Wait, in our problem, the angle at the satellite is \(20^{\circ}\). So \(\theta = 20^{\circ}=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Let minor arc be \(x\), major arc be \(360 - x\). So \(20=\frac{1}{2}((360 - x)-x)\)

\(20=\frac{1}{2}(360 - 2x)\)

\(40 = 360-2x\)

\(2x=360 - 40 = 320\)

\(x = 160^{\circ}\)? Wait, no, that's wrong. Wait, no, the angle between two tangents: the measure of the angle is equal to \(180^{\circ}-\) the measure of the central angle of the minor arc? No, let's check with a unit circle. If the minor arc is \(80^{\circ}\), major arc is \(280^{\circ}\), then the angle is \(\frac{1}{2}(280 - 80)=\frac{1}{2}(200) = 100^{\circ}\), which is not. Wait, I think I made a mistake. The correct formula for the angle formed by two tangents from an external point is: the measure of the angle is equal to \(180^{\circ}-\) the measure of the minor arc? No, no. Wait, the sum of the angle at the external point and the measure of the minor arc's central angle is \(180^{\circ}\)? No, that's not. Wait, let's look at the formula again. 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. The intercepted arcs are the major arc and the minor arc. So \(\theta=\frac{1}{2}(m\overset{\frown}{major}-m\overset{\frown}{minor})\). Since \(m\overset{\frown}{major}+m\overset{\frown}{minor}=360^{\circ}\), we can write \(m\overset{\frown}{major}=360 - m\overset{\frown}{minor}\). So \(\theta=\frac{1}{2}((360 - m\overset{\frown}{minor})-m\overset{\frown}{minor})=\frac{1}{2}(360 - 2m\overset{\frown}{minor}) = 180 - m\overset{\frown}{minor}\). So \(\theta=180 - m\overset{\frown}{minor}\), which means \(m\overset{\frown}{minor}=180-\theta\)? No, that can't be. Wait, if \(\theta = 20^{\circ}\), then \(m\overset{\frown}{minor}=180 - 20=160^{\circ}\)? No, that doesn't make sense. Wait, I think I had the formula reversed. The correct formula is \(\theta=\frac{1}{2}(m\overset{\frown}{minor}-m\overset{\frown}{major})\)? No, no. Let's refer to the standard formula: The measure of an angle formed by two tangents (or a tangent and a secant) drawn from an external point to a circle is half the difference of the measures of the intercepted arcs. The larger arc minus the smaller arc. So if the angle is \(20^{\circ}\), then \(20=\frac{1}{2}(L - S)\), where \(L\) is the major arc and \(S\) is the minor arc, and \(L + S=360\). So \(L=360 - S\). Substitute into the formula: \(20=\frac{1}{2}((360 - S)-S)= \frac{1}{2}(360 - 2S)=180 - S\). Then \(S=180 - 20 = 160\)? No, that's not right. Wait, maybe the diagram is a tangent - secant? No, the two lines from the satellite to the circle: if the circle is Earth, the two lines are tangents. Wait, maybe the angle is half the measure of the difference between the major and minor arcs. Wait, let's take a simple case: if the minor arc is \(100^{\circ}\), major arc is \(260^{\circ}\), then the angle is \(\frac{1}{2}(260 - 100)=80^{\circ}\). If the angle is \(20^{\circ}\), then \(\frac{1}{2}(L - S)=20\), so \(L - S = 40\), and \(L+S = 360\). Solving the system of equations: add the two equations \(L - S+L + S=40 + 360\), \(2L=400\), \(L = 200\), \(S=360 - 200 = 160\)? No, that's not. Wait, I think I made a mistake in the formula. The correct formula for the angle between two tangents is that the angle is equal to \(180^{\circ}-\) the measure of the central angle of the minor arc. Wait, no, let's check with a source. The measure of an angle formed by two tangents to a circle is equal to half the difference of the measures of the intercepted arcs. The intercepted arcs are the major arc and the minor arc. So \(\angle A=\frac{1}{2}(m\overset{\frown}{BCD}-m\overset{\frown}{BD})\) (where \(A\) is the external point, and \(BCD\) is the major arc, \(BD\) is the minor arc). So if \(\angle A = 20^{\circ}\), then \(20=\frac{1}{2}(m\overset{\frown}{major}-m\overset{\frown}{minor})\). Let \(m\overset{\frown}{minor}=x\), \(m\overset{\frown}{major}=360 - x\). Then \(20=\frac{1}{2}((360 - x)-x)\), \(40 = 360-2x\), \(2x=320\), \(x = 160\)? No, that's not. Wait, maybe the diagram is a tangent and a secant? No, the two lines from the satellite to the circle: both are tangents. Wait, maybe the angle is half the measure of the minor arc? No, that would be if it's an inscribed angle. Wait, no. Wait, let's think differently. The triangle formed by the center \(O\) and the two points of tangency and the satellite: the radius is perpendicular to the tangent at the point of tangency. So \(\angle O\) (the central angle) and the angle at the satellite are supplementary? No, the quadrilateral formed by the center, two points of tangency, and the satellite has two right angles (radius - tangent angles) and the angle at the satellite and the central angle. So the sum of the interior angles of a quadrilateral is \(360^{\circ}\). So \(90 + 90+\angle\text{satellite}+\angle O=360\), so \(\angle O + 20=180\), so \(\angle O = 160^{\circ}\). And the measure of the minor arc \(x\) is equal to the measure of the central angle \(\angle O\). So \(x = 160^{\circ}\)? Wait, no, the central angle corresponding to the minor arc is equal to the measure of the minor arc. So if the central angle is \(160^{\circ}\), the minor arc is \(160^{\circ}\). But let's check the options. The options are \(40^{\circ}\), \(80^{\circ}\), \(160^{\circ}\), \(320^{\circ}\). Wait, maybe I made a mistake in the formula. Let's start over.

The correct formula for the angle formed by two tangents from an external point: the measure of the angle is equal to \(180^{\circ}-\) the measure of the central angle of the minor arc. Wait, no, the central angle of the minor arc is equal to the measure of the minor arc. The angle at the external point (satellite) and the central angle of the minor arc: in the quadrilateral with vertices at the center \(O\), point of tangency \(A\), satellite \(S\), point of tangency \(B\), we have \(\angle OAS = 90^{\circ}\), \(\angle OBS=90^{\circ}\), \(\angle ASB = 20^{\circ}\), and \(\angle AOB=x\) (the central angle, equal to the measure of the minor arc). So \(90 + 90+20 + x=360\), so \(x=160^{\circ}\). Yes, that's correct. Because the sum of the interior angles of a quadrilateral is \(360^{\circ}\), and the two radius - tangent angles are \(90^{\circ}\) each, the angle at the satellite is \(20^{\circ}\), so the central angle (and thus the minor arc \(x\)) is \(360-(90 + 90+20)=160^{\circ}\).

Answer:

\(160^{\circ}\)