Question

a satellite views the earth at an angle of 20°. what is the arc measure, x, that the satellite can see?
○ 40°
○ 80°
○ 160°
○ 320°
diagram: circle labeled earth with center o, two tangent lines from point satellite to the circle, forming a 20° angle at satellite, arc x between the two tangent points on earth

Explanation:

Step1: Recall the tangent - radius property

The lines from the satellite to the Earth's surface are tangents to the Earth (a circle). The radius to a tangent point is perpendicular to the tangent line. So, if we connect the center of the Earth (O) to the two tangent points, we form two right angles (90° each) with the tangent lines.
Let the center of the Earth be O, and the two tangent points be A and B. Then, $\angle OAS=\angle OBS = 90^{\circ}$, where S is the satellite.

Step2: Use the sum of angles in a quadrilateral

In quadrilateral OASB, the sum of interior angles is $360^{\circ}$. We know $\angle ASB = 20^{\circ}$, $\angle OAS=\angle OBS = 90^{\circ}$. Let $\angle AOB=x$ (the central angle corresponding to arc AB, which is the arc the satellite can see).
So, $\angle OAS+\angle ASB+\angle OBS+\angle AOB = 360^{\circ}$
Substitute the known values: $90^{\circ}+20^{\circ}+90^{\circ}+x = 360^{\circ}$
Simplify the left - hand side: $200^{\circ}+x = 360^{\circ}$
Subtract $200^{\circ}$ from both sides: $x=360^{\circ}- 200^{\circ}=160^{\circ}$
Alternatively, we can use the property that the measure of the central angle is equal to $180^{\circ}-\theta$, where $\theta$ is the angle at the external point (the satellite) for a circle. Wait, no, the correct relation is that if we have an external angle $\theta$ (the angle at the satellite) between two tangents, the measure of the intercepted arc (the arc between the two tangent points) is $180^{\circ}-\theta$? No, wait, let's re - derive.
The triangle formed by the center O and the two tangent points A and B: OA and OB are radii, so OA = OB. The lines SA and SB are tangents, so SA = SB (tangents from a common external point to a circle are equal). So, triangle SAB is isosceles, and triangle OAS and OBS are right - angled.
The central angle $\angle AOB$ and the angle at S ($\angle ASB$) are related by $\angle AOB + \angle ASB=180^{\circ}$? No, wait, in the quadrilateral OASB, as we calculated before, $\angle AOB=180^{\circ}-\angle ASB$? Wait, no, our first calculation was:
$90 + 20+90+\angle AOB = 360$
$200+\angle AOB = 360$
$\angle AOB = 160^{\circ}$, which is correct. Because the sum of angles in a quadrilateral is $360^{\circ}$, and two of the angles are right angles (90 each) and one is $20^{\circ}$, so the fourth angle (the central angle) is $360-(90 + 20+90)=160^{\circ}$.

Answer:

$160^{\circ}$ (corresponding to the option "160°")