Q13 of 61 Page 215

A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height. The angle of elevation of the balloon from the boy at an instant is 60°. After 2 minutes, from the same point of observation, the angle of elevation reduces to 30°. If the speed of wind is 29√3 m/min. then, find the height of the balloon from the ground level.


Here, Distance covered by the balloon = BC


We know,


Distance = Time x Speed


⇒ BC = Time x Speed


= 2 x 29√3


= 58√3 m


Let AB = x


⇒ AC = x + 58√3


In triangle DAC,


∠DAC = 30°


We know,






Now, in triangle EAB,


∠EAB = 60°





⇒ EB = √3x


∵ EB = DC



⇒ x + 58√3 = 3x


⇒ 2x = 58√3



⇒ x = 29√3 m


And, Height of the balloon from ground level EB = √3 x


= 29 √3 (√3)


= 87 m


Hence height of the balloon from ground level is 87 m.


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