Question

plz tell the answer

Answer 1

*Hi there *

Here you go the answer I have got ...
But I am not sure about the answer..

But I can assure you that the process is correct...

Hope this helps you...

REGARDS...

Answer 2

Solution :

_____________________________________________________________

Given :

Condition :

  ⇒ The angle of elevation of an aircraft from a point on horizontal ground is found to be 30°.

  ⇒ The angle of elevation of same aircraft after 24 seconds which is moving horizontally to the ground is found to be 60°.

  ⇒ The height by which the air craft was flying vertically above the ground is 3600√3 m,.

________________________

From the above information ,

We can find that,.

(As per the given diagram),.

In Δ PCQ & Δ ABC,

⇒ ∠Q = ∠B = 90°

⇒ ∠ACB = 30°

⇒ ∠ PCQ = 60°

⇒ AP = BQ

⇒ PQ = AB = 3600√3 m

_____________________________________________________________

To Find :

The velocity of air craft :

_____________________________________________________________


In Δ CPQ

We know that,

∠ PCQ = 60°

∴ Tan C = Tan 60°

⇒ Tan C = $\frac{PQ}{CQ}$

⇒ Tan 60° = $\frac{3600 \sqrt{3}}{CQ}$

⇒ $\sqrt{3} = \frac{3600 \sqrt{3}}{CQ}$

⇒ $CQ \sqrt{3} = 3600 \sqrt{3}$

∴ CQ = 3600 m.

_____________________________________

In Δ ABC

⇒ Tan C = $\frac{AB}{BC} = \frac{AB}{BQ + CQ}$

⇒ Tan 30° = $\frac{3600 \sqrt{3}}{BQ + 3600}$

⇒ $\frac{1}{ \sqrt{3}} = \frac{3600 \sqrt{3}}{BQ + 3600}$

⇒ $(3600 \sqrt{3} )( \sqrt{3}) = BQ + 3600$

⇒ $3600(3) = BQ + 3600$

⇒ $10800 = BQ + 3600$

⇒ $10800 - 3600 = BQ$

⇒ $BQ = 7200$ m

____________________________

We know that,

⇒ AP = BQ

∴ AP = 7200 m

The distance traveled by air craft in 24 seconds = 7200 m

______________

Velocity = $\frac{Displacement}{Time}$

⇒ $\frac{7200}{24}$

⇒ $\frac{24*300}{24}$

⇒ 300 m/s,.

                   ∴  The velocity of aircraft = 300 m/s

_____________________________________________________________

                                      Hope it Helps !!

⇒ Mark as Brainliest,.