Aboard an orbital research station, Emek and Tara prepared a communication antenna to contact a distant satellite. On the station's guidance diagram, the satellite's direction formed a right triangle with a horizontal reference line. The satellite was positioned 24 kilometers horizontally from the reference point and 10 kilometers above it. Tara used the tangent relationship, tan θ = 10/24, and found that the antenna needed to point about 22.6 degrees above the horizontal. Before transmitting, they also had to check a solar panel extending 8 meters horizontally from the antenna. Its upper edge was 3 meters above the antenna's base. Emek calculated that a signal traveling at 22.6 degrees would be about 8 tan(22.6 degrees), or 3.33 meters, above the base after 8 meters. Since 3.33 meters was greater than the panel's 3-meter height, the signal path cleared it by about 0.33 meter. Tara set the antenna to 22.6 degrees, and the station established a safe, steady connection with the satellite.

Challenges

Challenge 1 of 5

Challenges

Challenge 1 of 4
Hint
For an angle measured above the horizontal, tangent compares the vertical side with the horizontal side.
The satellite is 24 kilometers horizontally from the reference point and 10 kilometers above it. Which equation correctly represents the tangent of the antenna angle theta?
tan theta = 24/10tan theta = 10/24tan theta = 10/26tan theta = 24/26
Correct. The opposite side is 10 kilometers and the adjacent side is 24 kilometers, so tan theta = 10/24.
Identify the side opposite the antenna angle and the side adjacent to it before forming the tangent ratio.
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Hint
The horizontal and vertical distances are the legs of a right triangle. Use the Pythagorean theorem.
The satellite is 24 kilometers horizontally from the reference point and 10 kilometers above it. What is the straight-line distance, in kilometers, from the reference point to the satellite?
Correct. The straight-line distance to the satellite is 26 kilometers.
Square the two perpendicular distances, add them, and then take the square root.
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Hint
Use the rise-to-run ratio 10/24 and apply it to a horizontal distance of 12 meters.
The signal path rises at the same rate as the line from the reference point to the satellite. At a horizontal distance of 12 meters from the antenna, approximately how high above the antenna's base would the signal be?
3 meters4 meters5 meters10 meters
Correct. Using the same slope, the signal rises 5 meters over a horizontal distance of 12 meters.
Use the constant ratio of vertical rise to horizontal distance from the satellite triangle.
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Uydu Antenini Hizalamak
TR
ARC_G11_GM_02_STORY_01
Uydu Antenini Hizalamak
Aligning the Satellite Antenna
EN
ARC_G11_GM_02_STORY_01
Aligning the Satellite Antenna