The angle of elevation of an aeroplane from the point on on the ground...
Problem: The angle of elevation of an aeroplane from a point on the ground is 60 degrees after a flight of 10 seconds. On the same height, the angle of elevation from another point becomes 30 degrees. If the aeroplane is flying at the speed of 720 kilometers, find the constant height at which the aeroplane is flying.
Solution:
Let us assume that the height of the aeroplane from the ground is h.
Step 1: Finding the distance travelled by the plane in 10 seconds.
We know that the plane is flying at a speed of 720 km/hr.
Therefore, the distance travelled in 10 seconds is:
720 km/hr = (720 x 1000) m/hr [Converting km/hr to m/hr]
= (720 x 1000) / 3600 m/s [Converting m/hr to m/s]
= 200 m/s
So, the distance travelled by the plane in 10 seconds is:
Distance = Speed x Time
= 200 m/s x 10 s
= 2000 m
Step 2: Finding the horizontal distance of the plane from the two points.
Let us assume that the distance of the plane from the first point is x.
Then, the distance of the plane from the second point is (2000 - x) [As the plane is flying at a constant height h]
Step 3: Finding the height of the plane.
We know that the angle of elevation from the first point is 60 degrees.
Therefore, we can write:
tan 60 = h/x
√3 = h/x
h = √3 x
Similarly, the angle of elevation from the second point is 30 degrees.
Therefore, we can write:
tan 30 = h/(2000 - x)
1/√3 = h/(2000 - x)
h = (2000 - x)/√3
Step 4: Equating the two expressions for height.
As the height of the plane is constant, we can equate the two expressions for height obtained in Step 3.
√3 x = (2000 - x)/√3
On solving, we get:
x = 500√3
h = 500
Therefore, the constant height at which the plane is flying is 500 meters.
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