Question

A plane flying horizontally 1000root3 m above the ground is observed at an angle of elevation 60 degree from a point on the ground after 10 seconds the angle of elevation at a point of observation changes to 30 degree find speed of the plane in metre per second

Answer 1

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Answer 2

Answer:

200 m/s

Step-by-step explanation:

Refer the attached figure

A plane flying horizontally 1000√3 m above the ground i.e. EB = DC = 1000√3 m

A plane is observed at an angle of elevation 60 degree from a point on the ground i.e.∠EAB = 60°

After 10 seconds the angle of elevation at a point of observation changes to 30 degree i.e. ∠DAC = 30°

Now we are supposed to find the speed of the plane

In ΔAEB

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan60^{\circ} = \frac{EB}{AB}$

$\sqrt{3}= \frac{1000\sqrt{3}}{AB}$

$AB= \frac{1000\sqrt{3}}{\sqrt{3}}$

$AB= 1000$

In ΔDAC

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan30^{\circ} = \frac{DC}{AC}$

$\frac{1}{\sqrt{3}}= \frac{1000\sqrt{3}}{AC}$

$AC= \frac{1000\sqrt{3}}{\frac{1}{\sqrt{3}}}$

$AC=3000$

So, Distance travelled in 10 seconds = AC - AB = 3000-1000=2000 m

Speed = $\frac{Distance}{Time}$

Speed = $\frac{2000}{10}$

Speed =200 m/s

Hence the speed of the plane is 200 m/s.