Question

The angle of elevation of an aeroplane from a point on the ground is by And
flight of 30sec, the angle of elevation becomes 30°.If the aeroplane is flying at a
constant height of 3000
$\sqrt{3}$
m, find the speed of the aeroplane.
$\sqrt{3}$
​

Answer 1

Answer:

Answer:

240 km per hour

Step-by-step explanation:

We need distance travelled by aeroplane for speed

\begin{lgathered}x = h \cot(30) - h \cot(60) \\ = h( \sqrt{3} - \frac{1}{ \sqrt{3} } ) \\ = 1000 \sqrt{3} (\sqrt{3} - \frac{1}{ \sqrt{3} }) \\ = 1000(3 - 1) \\ = 2000 \: m\end{lgathered}

x=hcot(30)−hcot(60)

=h(

3

−

3

1

)

=1000

3

(

3

−

3

1

)

=1000(3−1)

=2000m

So speed will be

v = \frac{2000}{30} = \frac{200}{3} \frac{m}{s} = \frac{200}{3} \times \frac{18}{5} kmph = 240kmphv=

30

2000

=

3

200

s

m

=

3

200

×

5

18

kmph=240kmph

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 ✌️