Question

A 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is ${60}^{\circ }$. After sometime, the angle of elevation reduces to ${30}^{\circ }$. Find the distance travelled by the balloon during the interval?

• A. 52√3m
• B. 48√3m
• C. 58√3m
• D. 65√2m

Answer 1

The correct option is C

58√3m

Let CD be the girl

Let initial position of balloon A shift to another position B after sometime

In ΔACE, Tan ${60}^{\circ }$ =$\frac{AC}{CE}$

Tan ${60}^{\circ }$ = $\frac{(AF-EF)}{CE}$

$\sqrt{3}$= $\frac{(88.2-1.2)}{CE}$

CE = = 29$\sqrt{3}$m

In ΔBCG, Tan ${30}^{\circ }$ = $\frac{BG}{CG}$

Tan ${30}^{\circ }$ = $\frac{(BH-GH)}{CG}$

CG = 87$\sqrt{3}$m

Distance travelled by the balloon, EG = CG – CE = (87$\sqrt{3}$ - 29$\sqrt{3}$) = 58$\sqrt{3}$ m