Question

The angle of elevation of an airplane from a point on the ground is 60°. After a flight of
30 seconds, the angle of elevation becomes 30'. If the airplane is flying at a constant
height of 3000root3 m, find the speed of the airplane,

Answer 1

Answer:

Brainly.in

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Secondary School Math 5 points

The angle of elevation of an aeroplane from a point on the ground is 60 degree after a flight of 30 seconds the angle of elevation changes to 30 degree if the plane is flying at a constant height of 3600 under root 3 M find the speed of the plane in kilometre per hour

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Answers

OmGupta11

OmGupta11 Expert

In triangle ACE,

\tan(30) = \frac{ac}{ce} \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{ce} \\ ce = 3600 \sqrt{3} \times \sqrt{3} \\ ce = 10800 \: m

CE = 10800 m

AC = BD =

3600 \sqrt{3} \: m

In triangle BED,

\tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

speed = \frac{distance}{time} = \frac{7200}{30} = 240 \: m {s}^{ - 1}

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h

Answer 2

Answer:

${ \purple{ \tan 60° = \frac{3000 \sqrt{3} }{x} }}$

${ \purple{x= \frac{3000 \sqrt{3} }{ \sqrt{3} } = 3000m}}$

${ \purple{ \tan 30° = \frac{3000 \sqrt{3} }{ \frac{1}{ \sqrt{3} } } = 9000m}}$

Distance covered :

${ \purple{ 9000 - 3000 = 6000m}}$

${ \purple{speed = \frac{6000}{30} }}$

${ \huge{ \boxed{ \purple{speed = 200 m/s}}}}$

Step-by-step explanation:

Hope this helps you ✌️