A man from the top of a 100 metre high tower sees a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. The distance (in metres) travelled by the car during this time is

$an 60^{circ}=frac{mathrm{AB}}{mathrm{BD}}=frac{sqrt{3}}{1}$

$an 30^{circ}=frac{mathrm{AB}}{mathrm{BC}}=frac{1}{sqrt{3}}$

$frac{sqrt{3}}{mathrm{BC}}=frac{1}{sqrt{3}}$

BC = 3 unit

∴ DC = BC – BD

= 3 – 1 = 2 unit

According to question,

$mathrm{AB}=sqrt{3}=100$ metre

$1=frac{100}{sqrt{3}}$

2 unit $=frac{100}{sqrt{3}} imes 2=frac{200}{sqrt{3}} imes frac{sqrt{3}}{sqrt{3}}=frac{200 sqrt{3}}{3}$ metre