An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45⁰. After 20 seconds, the angle of elevation was observed to be 30⁰. Determine the height at which the aircraft flying. (use √3 = 1.732)

Solution:

It is given that

Speed of aircraft = 360 km/h

Distance covered by the aircraft in 20 seconds = (360 × 20)/ (60 × 60) = 2 km

Consider E as the fixed point on the ground

CD as the position of AB in height of aircraft

Take AB = CD = h km

In right triangle ARB

tan θ = AB/ EB

Substituting the values

tan 45⁰ = h/EB

1 = h/EB

EB = h

Here

ED = EB + BD = h + 2 km

In right triangle CED

tan 30⁰ = CD/ED

Substituting the values

1/√3 = h/ (h + 2)

√3h = h + 2

1.732h – h = 2

0.732h = 2

We know that 2km = 2000 m

h = 2000/0.732

h = (2000 × 1000)/ 732 = 2732 m

Chapter Test