Question

An aeroplane is flying at a constant height of 1200√3 meters. The angle of elevation of that aeroplane from a point on the ground is 60°. After a flight of 25 seconds, the angle of elevation changes to 30°. Find the speed of aeroplane.

Answer 1

Given:

$\text{The plane is at a height of }1200\sqrt{3} \text{ m high.}$

$\text{Angle of elevation is } 60^\circ$

$\text{Angle of elevation is } 30^\circ \text{ after 25 seconds}$

To Find:

$\text{Speed of the plane}$

Formulae Needed:

$\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

$\text{Speed} = \text{Distance} \div \text{Time}$

Solution

Find the ground distance from the point to the plane initially:

Let the distance be D₁

$\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

$\tan (60) = \dfrac{1200\sqrt{3} }{D_1}$

$D_1= \dfrac{1200\sqrt{3} }{\tan (60) }$

$D_1 = 1200 \text{ m}$

Find the ground distance from the point to the plane after 25 seconds:

Let the distance be D₂

$\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

$\tan (30) = \dfrac{1200\sqrt{3} }{D_2}$

$D_2= \dfrac{1200\sqrt{3} }{\tan (30) }$

$D_2 = 3600 \text{ m}$

Find the difference in distance:

$\text{Difference in Distance } = D_2 - D_1$

$\text{Difference in Distance } = 3600 - 1200$

$\text{Difference in Distance } = 2400 \text { m}$

$\text{Difference in Distance } = 2.4 \text{ km}$

Find the speed of the aeroplane:

$\text {Distance} = 2.4 \text { km}$

$\text {Time} = 25 \text { seconds}$

$\text {Time} = 25 \div 60 \div 60 \text { hour}$

$\text {Time} = \dfrac{1}{144} \text { hour}$

$\text{Speed} = \text{Distance} \div \text{Time}$

$\text{Speed} =2.4 \div \dfrac{1}{144}$

$\text{Speed} = 2.4 \times 144$

$\text{Speed} = 345.6 \text { km/h}$

$\boxed {\textbf {Answer : Speed = 345.6 km/h} }$

Answer 2

Step-by-step explanation:

Given An aeroplane is flying at a constant height of 1200√3 meters. The angle of elevation of that aeroplane from a point on the ground is 60°. After a flight of 25 seconds, the angle of elevation changes to 30°. Find the speed of aeroplane.

• So let the constant height be PQ    
• So let speed = m
• So RS = distance = speed x time
• RS = m x 25
•     = 25 m
• Consider triangle PQR
• So tan 60 = PQ / QR
• So √3 = 1200 √3 / QR
• QR √3 = 1200 √3
• QR = 1200
• Now consider triangle PQS
• So tan 30 = PQ / QR
• 1 / √3 = 1200 √3 /QR + RS
•        = 1200 x √3 /1200 + 25 m
• 1/√3 x 1200 + 25 m = 1200 √3
• 1200 + 25 m = 1200 x √3 x √3
• 1200 + 25 m = 1200 x 3
• 1200 + 25 m = 3600
• Or 25 m = 2400
• Or m = 2400 / 25
• Or m = 96

Reference link will be

https://brainly.in/question/15002366