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A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given. Use the Law of Sines to solve the problem. You must solve for BC first. Solve this problem in order.
Right triangle ADC has right angle D. B is between A and D. Point A is labeled Tower 1. Point B is labeled Tower 2. The distance between A and B is 7600 feet. Angle CAD measures 16 degrees. Angle CBD measures 24 degrees. Angle D is marked with a small square.
Find BC, the distance from Tower 2 to the plane, to the nearest foot.
Find CD the height of the plane from the ground, to the nearest foot.
First, we will solve for BC:
Let x be the distance BC.
Using the Law of Sines in triangle ABC:
sin(16°)/7600 = sin(24°)/x
Cross multiply:
x * sin(16°) = 7600 * sin(24°)
x = (7600 * sin(24°)) / sin(16°)
x ≈ 10388 feet
Next, we will solve for CD:
Using the Law of Sines in triangle ADC:
sin(90°)/x = sin(16°)/7600
Cross multiply:
x * sin(16°) = 7600 * sin(90°)
x = 7600 * sin(90°) / sin(16°)
x ≈ 2731 feet
Therefore, BC ≈ 10388 feet and CD ≈ 2731 feet.