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"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 12 degrees from 10:01 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)." WHOEVER ANSWERS THIS FIRST WILL GET BRAINLIEST!

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nooris0663

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

Now can you mark me brainliest please?

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

Now you can give ''nooris0663'' brainliest. :)

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