# 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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