Flight 4358 took off from the DFW airport at an elevation of 48°. If the airplane has been traveling for 1200 ft., approximately how far above the ground is the airplane?

The airplane is approximately 891.8 ft above the ground

The situation can be modelled with a right angle triangle.  The airplane has been travelling for 1200 ft. The angle of elevation is 48°.

How far above the ground is the height of the right angle triangle. Therefore

using trigonometric ratio one can find the height.

hypotenuse = 1200 ft

angle = 48°

height = ?

sin 48 = opposite / hypotenuse

sin 48° = h / 1200

cross multiply

1200 sin 48 = h

h = 891.773790573

h ≈ 891.8 ft

Therefore, the airplane is approximately 891.8 ft above the ground

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891.77 feet

Step-by-step explanation:

since it is not stated, i'm going to assume that 1200 feet is the actual distance the plane travels along the 48° elevation. (i.e its the hypotenuse)

(see attached sketch)

we can see that the elements of the diagram form a right triangle, this means that we can use trigonometry to solve for the elevation x,

sin (48°) = x / 1,200

x = 1,200 sin (48°)

x = 891.77 feet