Question

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The angle of elevation of an airplane from point A on the ground is 60°.
After a flight of 10 seconds, on the same height, the angle of elevation
from point A becomes 30°. If the airplane is flying at the speed of
720 km/hr, find the constant height at which the airplane is flying.​

Answer 1

Answer:

Let P and Q be the two positions of the plane and A be the point of observation. Let ABC be the horizontal

line through A. It is given that angles of elevation of the plane in two positions P and Q from a point A are

60° and 30° respectively.

So, ∠PAB = 60°, ∠QAC = 30°

Again given that PB = 3600√3 m

In ΔABP, we have

tan 60° = BP/AB

=> √3 = 3600√3/AB

=> AB = 3600 m

In ΔACQ, we have

tan 30° = CQ/AC

=> 1/√3 = 3600√3/AC

=> AC = 3600√3 * √3 = 3600 * 3 = 10800 m

Now, Distance = BC = AC – AB = 10800 m – 3600 m = 7200 m

Thus, the plane travels 3000 m in 30 seconds

Hence, speed of plane = 7200/30 m/sec = 240 m/sec

= (240/1000) * 60 * 60 km/hr

[Since 1 km = 1000 m, 1 hr = 60 minutes = 60 * 60 seconds]

= 864000/1000 km/hr

= 864 km/hr

Step-by-step explanation:

Answer 2

Answer:

√3 km

Step-by-step explanation:

Find the distance traveled in the 10 seconds:

Given speed = 720 km/h

1 hour = 720 km

1 sec = 720 ÷ 60 ÷ 60  = 0.2 km

10 sec = 0.2 x 10 = 2 km

Define x:

Let the distance between point A and the plane be x km

The distance between point A and the plane after 10 sec = (x + 2) km

Find the height of the plane at point A:

Let the height be H

tan(60) = H/x

H = x tan(60)

Find the height of the plane at point A after 10 sec:

tan(30) = H/(x + 2)

H = (x + 2) tan(30)

Solve x:

Since both the height are equal

x tan(60) = (x + 2) tan(30)

x tan (60) = x tan (30) + 2 tan (30)

x tan (60) - x tan (30) = 2 tan (30)

x( tan (60) - tan (30) ) = 2 tan (30)

x = 2 tan(30) / (tan(60) - tan(30) )

x = 1 m

Find the height:

H = x tan(60)

H = tan(60)

H = √3 km ( or 1.73 km)

Answer: √3 km