Question

The angle of elevation of an aeroplane from
a point P on the ground is 60°. After 12
seconds, the angle of elevation changes to 30°
If the plane is flying horizontally at a speed
of 600/3 km/h, find the height at which it is
flying​

Answer 1

★Given:-

• The angle of elevation of an aeroplane from  a point P on the ground is 60°.
• After 12  seconds, the angle of elevation changes to 30°
• Plane is flying horizontally at a speed  of 600/3 km/h

★To find:-

• The height at which it is flying.

★Solution:-

Let,

• A=position of plane when the angle of elevation is 60°.
• D = position of plane when angle of elevation is 30°.
• AB = h ; DC = h

We have:

• Speed = 600√3 km/hr
• Time =  1/300 hr

(By converting units of time,

12 sec = 12/3600

          =1/300 hr)

Using the formula,

✦Distance = speed×time

Putting values,

⇒ Distance =(600√3)× 1/300

                   =2√3 km

Therefore,

→AD = 2√3 km

Using the formula,

✦Tanθ = Opposite side/adjacent side

From ΔAPB,

→Tan 60°=AB/PB

(Tan 60° = √3)

⇒ √3 = h/PB

⇒ PB = h/√3

→Tan 30°=DC/PC

(Tan 30° = 1/√3)

⇒ 1/√3 = h/PC+BC -------(1)

We have AD = 2√3 km

As, ABCD is a rectangle and opposite sides of a rectangle are equal,

→BC = 2√3 km

Now substituting the values of PB & BC in equation (1),

⇒ 1/√3 = h/(h/√3 + 2√3)

⇒ 1/√3 = h/(h+6/√3)

⇒ h = 6+h/3

⇒ 3h = 6+h

⇒ 2h = 6

⇒ h = 6/2

⇒ h = 3km

Hence,

The plane is flying at a height of 3km.

______________

Answer 2

Answer:

Correct Question:-

1. The angle of elevation of a jet plane from a point on the ground is 60°. After a flight of 30 second the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 3600√3 m. Find the speed of jet plane.

To find,

The speed of jet plane

Formula used,

$speed = \frac{distance}{time}$

Solution:-

In ∆ ABD

$Tan \: 60°= \frac{BD}{AB} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{AB } \\ \sqrt{3} AB=3600\sqrt{3} \\ AB= 3600 \: m$

Now in ∆ ACE

$Tan \: 30°= \frac{CE}{AC} \\ \frac{1}{ \sqrt{3} } = \frac{CE}{AC } \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{AC } \\ AC=3600 \sqrt{3} . \sqrt{3} A \\ = 3600×3 \\ AC=10800 \: m \\ ➪DE=BC=AC-AB \\ 10800-3600 \\ \\ ⟹DE=7200 \: m \\ ⟹Speed \: of \: jet \: plane= \frac{Distance}{Time} \\ = \frac{7200}{30} \\ ⟹240 \: m/s= \frac{240×18}{5} =864 \: km$

So the required final answer is 864 km