Math

The angle of elevation from Jet fighter. A ground on 60 degree after a flight takeoff 15 seconds the angle of elevation 30 degree if you at speed of 720 km per hour then find the constant height reached at the flying

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  1. I will assume the following:
    when t = 0 , angle of elevation is 60°
    when t = 15 seconds, angle of elevation is 30°
    horizontal speed is 720 km/h

    let the constant height that the plane is flying be h m

    I have a right-angled triangle with height h m, base of x m
    and base angle of Ø° degrees

    speed of 720 km/h = 720000/3600 m/s = 200 m/s

    tanØ = h/x -------> h = x tanØ

    case 1: when Ø = 60° , tan 60° = √3 , height = h, base = x
    h = x√3

    case 2: when Ø = 30° , horizontal distance = x + 15(200) = x + 3000
    tan30° = 1/√3
    h = (x + 3000)(1/√3) = (x+3000)/√3

    √(x+3000)/√3 = x√3
    x+3000 = 3x
    x = 1500

    then in : h = x√3
    h = 1500√3 or appr 2598 m or 2.598 km

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  2. typo in 5th last line:
    should be
    (x+3000)/√3 = x√3 , not √(x+3000)/√3 = x√3

    has no effect on what follows

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  3. 750 km/hr = 200 m/s
    so, in 15 seconds, the plane has gone 3000m
    Draw a diagram. It should be clear that if the plane's height is h, then
    h cot30° - h cot60° = 3000
    h = 3000/(√3 - 1/√3) = 1500√3 meters

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