The boy whose eye level is 1 3m from the ground, spots a balloon moving with the wind in a - Maths - Some Applications of Trigonometry

Question

The boy whose eye level is 1 3m from the ground, spots a
balloon moving with the wind in a horizontal line at some height
from the ground The angle of elevation of the balloon from the eyes
of the boy at an instant is 600 After 2 seconds, the
angle of elevation reduces to 300 If the speed of the
wind is 29√3m/s,then find the height of the balloon from the
ground
Hint : Distance covered by the balloon in 2sec = 29√3 x
2 = 58√3m

Ajanta Trivedi · Accepted Answer

let the boy spot the balloon at point A and the angle of elevetion from his eye point P to the balloon is 60 deg PQ=1 3 m let after 2 seconds the balloon reaches to the point B and now its angle of elevation from the eye is 30 deg QCF is the horizontal line at the ground level since speed of the wind = $29 \sqrt{3} m / \sec$ the distance covered by the balloon in 2 sec be AB= $29 \sqrt{3} \times 2 = 58 \sqrt{3} m$ (a) in the right angled triangle APD, $\tan 60 = \frac{A D}{P D} \sqrt{3} = \frac{A D}{P D} \Rightarrow P D = \frac{1}{\sqrt{3}} A D m$ (1) in the right angled triangle BEP, $\tan 30 = \frac{B E}{P E} \frac{1}{\sqrt{3}} = \frac{A D}{P E} [\sin c e B E = A D] P E = \sqrt{3} A D m (2)$ PE=PD+DE PE=PD+AB [since DE=AB] $P E = P D + 58 \sqrt{3}$ [using (a)] from (1) and (2), substitute the values in terms of AD $\sqrt{3} A D = \frac{1}{\sqrt{3}} A D + 58 \sqrt{3} 3 A D = A D + 174 2 A D = 174 A D = \frac{174}{2} = 87 m$ therefore the height of the balloon from the ground level is AC=AD+DC AC=AD+PQ AC=87+1 3=88 3 m thus the height of the balloon from the ground is 88 3 m