Question

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After a flight of 10 seconds, it's angle of elevation is 30° from the same point on the ground. Find the speed of the aeroplane in km/hour

Answer 1

Answer:

Step-by-step explanation:

tan60°= 1/d =√3

⇒d =1/√3 km 

now, the aeroplane cover (d+x) km

tan30°= 1/(d+x)=1/√3

⇒(d+x) = √3

⇒x = √3 -(1/√3) = (3-1)/√3 =2/√3= (2×1.732)/3 = 1.15 km

so speed =distance/time = 1.15/(10/3600) =414 km/hr

Answer 2

ANSWER:

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Let A and B be the first and second position of the plane respectively.

Let P be the point of observer.

Ang.APD and ang.BPC are the angles of elevation.

Ang.APD=60° and

Ang. BPC=30°

AD and BC represent constant height at which the plane is flying.

Therefore,

AD=BC=100m.

In right angled ADP,

tan 60°=AD/PD.

$\\ = > \sqrt{3} = \frac{1000}{PD}$

Therefore,

$\\ = > PD = \frac{1000}{ \sqrt{3} }$

$\\ = > PD = \frac{1000 \sqrt{3} }{3}$

$\\ = > PD = 577.367m$

In right angled BCP,

tan 30°=BC/PC

$\\ = > \frac{1}{ \sqrt{3} } = \frac{1000}{PC}$

$\\ = > PC = 1000 \sqrt{3}$

$\\ = > PC = 1732m$

$\\ = > PD + DC = PC........(P - D - C)$

$\\ = > 577.367 + DC = 1732$

$\\ = > DC = 1732 - 577.367$

$\\ = > DC = 1154.633m$

⬛️ ABCD is a rectangle.

$\\ = > AB = DC = 1154.633m.......(opposite \: sides \: of \: rectangle)$

Therefore,

Distance covered in 10 second is 1154.633m.

Therefore,

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

$\\ speed = \frac{AB}{10sec}$

$\\ speed = \frac{1154.633m}{10secs \: }$

$\\ speed = \frac{1154.633}{1000} \div \frac{10}{3600}......(1km = 1000m \: and \: 1hr = 3600 \: seconds)$

$\\ speed = \frac{1154.633}{1000} \div \frac{3600}{10}$

$\\ speed = 415.667kmphr.$

Therefore,

Speed of the plane =415.667km/hr.