Question

a man on the top of a vertical tower observes a car moving at a uniform speed towards nim. if he takes 12 min for the angle of depresion to change from30°to45°. how soon after this the car will reach tower

Answer 1

Let AB is a tower, car is at point D at 30°

and goes to C at 45° in 12 minutes.

In ∆ ABC,

AB

BC = tan 45°

⇒

h

x = 1 ⇒ h = x ...(i)

In ∆ ABD,

AB

BD

= tan 30°

⇒

h

x y + = 1

3

⇒ h = x y +

3 ...(ii)

Comparing eq. (i) & (ii), we get

x = x y +

3

⇒ 3 x = x + y

⇒ ( ) 3 1 − x = y

Car covers the distance y in time = 12 min

So ( ) 3 1 − x distance covers in 12 min

Distance x covers in time = 12

3 1

3 1

− 3 1

×

+

+

=

12 3 1

3 1 2

( ) +

− =

= 6 ( ) 3 1 + min

= 6 × 2.732 = 16.39

Now car reaches to tower in 16.39 minutes. Ans