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The pilot of an aircraft flying horizontally at a speed of 1200km/hr. Observes that the angle of depression of a point on the ground changes from $30 °$ to $45 °$ in 15 seconds. Find the height at which the aircraft is flying.

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7.83 km

6.83 km

8.83 km

9.83 km

answer is A.

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Detailed Solution

Given that the pilot of an aircraft flying horizontally at a speed of 1200km/hr.
We have to find the height at which the aircraft is flying.
The value for the trigonometric identities can be evaluated by using the right-angled triangle the value for

tan θ

can be expressed as

tan θ = opposite adjacent

Since the speed of an aircraft is

1200 km / hr

and, after 15 sec, the angle of depression changes from

30 °

45 ° .

Therefore,

A B = 1200 × 1000 × 15 sec 3600 sec ⇒ A B = 5000 m

Assume the height from ground be H and horizontal distance be X.
Then, by using the figure geometry,
Question Image

In triangle B D M tan θ = B D D M ⇒ tan 45 ° = H D M ⇒ 1 = H D M ⇒ D M = H

In triangle A C M ⇒ tan θ = A C C M ⇒ 13 = H C D + D M ⇒ 13 = H 5000 + H ⇒ H 3 = 5000 + H

On further solving the above expression,

⇒ H = 5000 3 − 1 ⋅ 3 + 1 3 + 1 ⇒ H = 5000 (1732) + 1 3 − 1 ⇒ H = 6830 m      [1km=1000m] ⇒ H = 6.83 km

The height at which the aircraft is flying is 6.83 km.
Therefore, the correct option is 1.

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