Sangeetha Pulapaka
0

Let the point of observation be A. 3 minutes is the time for the angle of depression to change from 30 ^\circ to 60^\circ . So the distance between CD is 3v if speed of car = v metre/minute. Let the time taken to reach from D from B is t minutes. Distance between BD = vt and distance between DC = 3v.

\angle ACB= 30^\circ

In \Delta ABC

tan 30^\circ = \frac{AB}{BC}

\frac{1}{\sqrt{3}} = \frac{AB}{vt+3v}

AB =  \frac{ vt+3v}{\sqrt{3}}

In \Delta ABD

Tan60^\circ = \frac{AB}{BD}

\sqrt{3} = \frac{AB}{VT}

\sqrt{3}vt = AB

\sqrt{3}vt = \frac{3v+vt}{\sqrt{3}}

2t = 3

t = 3/2 = 1.5 minutes