Question

The angle of elevation of a jetplane from a point A on the ground is 45'. After a flight of 10 seconds,the angle elevation changes to 60' If the jet plane is flying at a constant height 3km draw a suitable diagram to find the speed of the plane ​

Answer 1

The speed of the plane is 790.6 km/hr.

Step-by-step explanation:

See the attached diagram.

Let the height of the jet plane is AB = 3 km.

So, the horizontal distance of the plane from point A will also be 3 km.

Then, AB = BC = 3 km.

{As the angle of elevation is 45° and tan 45° = 1}

Now, from the right triangle Δ  ABD,

$\tan 60^{\circ} = \frac{BD}{AB} = \frac{BD}{3}$

⇒ BD = 3 tan 60° = 5.196 km

Now, CD = BD - BC = 5.196 - 3 = 2.196 Km.

Now, the plane moves from C to D in 10 seconds.

Therefore, the speed of the plane is $\frac{2.196}{10} \times 3600 = 790.6$ km per hour. (Answer)