Question

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any distant is 600 After some time, the angle of elevation reduces to 30° . Find the distance travelled by the balloon during the interval​

Answer 1

Here,

AC=DE=(88.2-1.2)m=87m

In ∆ABC

Perpendicular (P)=AC,Base(b)=BC

We know

$\rm tan \theta = \frac{P}{B}$

Here,

$\rm tan60 = \frac{AC}{BC}$

$\rm \sqrt{3} = \frac{87}{BC} \\ \implies \boxed{ \rm \: BC = \frac{87}{ \sqrt{3} } }$

In ∆BDE

perpendicular (p)=DE, Base(b)=BE

We know

$\rm tan \theta = \frac{P}{B}$

Here,

$\rm tan30 = \frac{DE}{BE}$

$\rm \frac{1}{ \sqrt{3} } = \frac{87}{BE} \\ \implies \boxed{ \rm BE = 87 \sqrt{3}}$

CE=BE-BC

$\rm \: CE = 87 \sqrt{3} - \frac{87}{ \sqrt{3} } \\ \implies \rm \: CE = \frac{174 \times \sqrt{3} }{3} = 58 \sqrt{3}$

So distance travelled by balloon is 58√3