Question

Challenge for Brainlians:

A boy spots a balloon moving with wind in a horizontal plane at a height of 88.2 m from the ground level. The distance of his eye level from the ground is 1.2m. The angle of elevation of the ball in from his eyes at an instant is 60°. After some time, from the same point of observation, the angle of elevation of the balloon reduces to 30°. Find the distance covered by the balloon during the interval.

Answer if you can.​

Answer 1

Bro, Kindly refer the attachment.

Let A be the point of observation of the boy.

Let D and E be the position of the balloon when it's angles of elevation are 30° and 60° respectively.

Let B and C be the points on horizontal line such that BE = CD. M, N and O be the points on ground such that

AM = NB = CO = 1.2 m

Given angles are EAB = 60° and DAC = 30°

Given sides are AM = OC = 1.2 m

and OD = 88.2 m

Also, we know that BE = CD = 87 m since BE is parallel to CD.

In ∆EAB, B is a right angle .

tan 60° = BE/ AB ==> AB = 87/ tan 60°

===> 87/√3 = 29√3

AB = 29√3 m

In right angle triangle ∆DAC,

tan 30° = DC/AC

AC = 87/tan 30°

===> 87√3

Therefore the distance covered by the balloon is ED = BC = AC - AB

====> 87√3 - 29√3

=====> 58√3 m.

=====> 100.456 m

Therefore , the distance covered by the balloon is 100.46 m