Question

A straight highway leads to the foot of a tower. A man standing at the
top of the tower observes a car at an angle of depression of 30, which is
approaching to the foot of the tower with a uniform speed. Six seconds later, the
angle of depression of the car is found to be 60°. Find the time taken by the car
to reach the foot of the tower from this point. [CBSE 2009) ​

Answer 1

$\huge\bf\underline{answer}$

Let AB is the tower and BD is the highway.

Now from triangle ADB,

tan30 = AB/BD

=> 1/√3 = AB/BD

=> AB = BD/√3 .............1

Again from triangle ACB

tan60 = AB/BC

=> √3 = AB/BC

=> AB = BC√3 ........2

from equation 1 and 2

BD/√3 = BC√3

=> (BC + CD)/√3 = BC√3

=> BC + CD = BC√3*√3

=> BC + CD = 3BC

=> 3BC - BC = CD

=> 2BC = CD

=> BC = CD/2

Since time taken by car to cover CD = 6 Second

So time taken by car to cover BC = 6/2 = 3 seconds

$\huge\bf\underline{hope \: it \: helps \: u}$

Answer 2

let CD=h

angle of depression=30

after 6 sec angle of depression=60

so let AB=y and BC=x

in triangle BCD

tan60=CD/BC

$\sqrt{3}$ =h/x

h=$\sqrt{3}$ x....(i)

in triangle ACD

tan30=CD/AC

1/$\sqrt{3}$ =h/x+y

x+y= $\sqrt{3}$ h

from eq(i)

x+y=($\sqrt{3}$)$\sqrt{3}$

x+y=3x....(ii)

It is given that a car moves from point A to B in 6 sec.

let speed =k km/s

time=distance/speed

6=y/k

y=6k

on putting y=6k in eq(ii)

x+6k=3x

6k=2x

x=3k

time=distance/speed

=x/k

=3k/k

=3 seconds