Question

The angle of elevation of an airplane from a point on the ground is 600. After a flight of
30 seconds, the angle of elevation becomes 300
.If the airplane is flying at a constant height of 3000√3 m, find the speed of the airplane.​

Answer 1

ANSWER:-

Given:

The angle of elevation of an airplane from a point A 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 3000√3m.

To find:

The speed of the airplane.

Solution:

We have,

•angle BAC=60° &

•angle DAE=30°

•It is also given that Height of the airplane is CB= 3000√3m.

Since,

The airplane is flying at constant height,

CB= ED= 3000√3m.

Therefore,

In right ∆ABC,

[tan60° = √3]

$tan60 \degree = \frac{BC}{AB} \\ \\ = > \sqrt{3} = \frac{3000 \sqrt{3} }{AB} \\ \\ = > \sqrt{3} ab = 3000 \sqrt{3} \\ \\ = > AB = \frac{3000 \sqrt{3} }{ \sqrt{3} } \\ \\ = > AB = 3000m..................(1)$

&

In right ∆ADE,

[tan30°= 1/√3]

$= > tan30 \degree = \frac{DE}{AD} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{3000 \sqrt{3} }{AB + BD} \\ \\ = > AB + BD = 3000 \sqrt{3} \times \sqrt{3} \\ \\ = > AB + BD = 3000 \times 3 \\ \\ = > AB + BD = 9000m........................(2)$

Therefore,

Putting the value of equation (1) in equation (2), we get;

=) 3000m + BD= 9000m

=) BD= (9000-3000)m

=) BD= 6000m

Now,

Speed of airplane:

=) 6000/30

=) 200 m/s

Hence,

The speed of the airplane is 200m/s.

Hope it helps ☺️

Answer 2

Given:-

• The angle of elevation of an aeroplane from a point on the ground is 600.
• The angle of elevation becomes 300 after a flight of 30 sec.
• Aeroplane is flying at a constant height of 3000√3 m.

Find:-

Speed of the aeroplane.

Solution:-

*Refer the attachment for figure.

In ΔABE

=> tan 30° = $\frac{BE}{AC}$

=> $\frac{1}{\sqrt{3}}$ = $\frac{3000\sqrt{3}}{AB\:+\:BC}$

Cross-multiply them

=> $3000( \sqrt{3} )(\sqrt{3} )$ = $AB\:+\:BC$

=> $AB\:+\:BC$ = $3000(3)$

=> $AB\:+\:BC$ = $9000$ ___ (eq 1)

Similarly,

In ΔACD

=> tan 60° = $\frac{DC}{AB}$

=> tan 60° = $\frac{DC}{AB}$

=> $\sqrt{3}$ = $\frac{3000\sqrt{3}}{AB}$

Cross-multiply them

=> $AB\:=\:\frac{3000\sqrt{3}}{\sqrt{3}}$

=> $AB\:=\:3000$

Substitute value of AB in (eq 1)

=> $3000\:+\:BC$ = $9000$

=> $BC\:=\:6000$

Speed = $\bold{\frac{6000}{30}}$

=> 200 m/s

•°• Speed of aeroplane is 200 m/s.

In km/hr.

=> $\frac{200\:\times\:3600}{1000}$

=> $2\:\times\:360$

=> $720$

•°• 720 km/hr is the speed of the aeroplane.