The height of the airplane is 335.17 ft.
Let PQ be the base, AB be the vertical distance from the ground to the aircraft.
One observer is standing at P and the angle of inclination from P to A is 50°.
Another observer is standing at Q and the angle of inclination from Q to A is 25°.
We know that, PB + BQ = 1000 ft
PB = 1000 - BQ -----(1)
When we consider a triangle PAB, Tan 50° = AB/PB
AB = PB Tan 50° ------(2)
Consider the triangle ABQ, Tan 25° = AB/BQ
AB = BQ Tan 25° ------(3)
Equating (2) and (3) we have,
PB Tan 50° = BQ Tan 25° -----(4)
Substitute (1) in (4), we get,
(1000 - BQ) Tan 50° = BQ Tan 25°
1000 Tan 50° - BQ Tan 50° = BQ Tan 25°
BQ (Tan 25° + Tan 50°) = 1000 Tan 50°
BQ = 1000 Tan 50°/ (Tan 25° + Tan 50°)
BQ = 718.76 ft
PB = 1000 - 718.76 = 281.24 ft
AB = PB Tan 50° = 281.24* Tan 50° = 335.17 ft
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