A test rocket is launched at 10 am. By 10:01 am, the rocket has reached an altitude of 15,000 feet. From point A, on ground level 25,000 feet from the launch pad, it is observed that the angle of elevation increases by 120 from 10:01 am to 10:02 am. Find the altitude of the rocket at 10:02 am to the nearest tenth of a foot. (Assume that the rocket's path is vertical.)
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Answer:
23283.33 foot
Step-by-step explanation:
Tan θ = Perpendicular Base
=> Tanθ = 15000/25000
=> Tanθ = 3/5
=> Tanθ = 0.6
=> θ = 30.96°
Tan(θ + 12) = H/25000
=> H = 25000 * Tan(θ + 12)
Tan(θ + 12) = Tan(30.96° + 12)
=> Tan(θ + 12) = Tan(42.96°)
=> Tan(θ + 12) = 0.931333
=> H = 25000 * 0.931333
=> H = 23283.33
the altitude of the rocket at 10:02 am to the nearest tenth of a foot = 23283.33 foot
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