Question

A man on the top of a vertical observation tower observes a car moving at a uniform speed

coming directly towards it. If it takes 12 minutes for the angle of depression to change from
30°
to 45°
, how long will the car take to reach the observation tower from this point?
​

Answer 1

$\huge \boxed{ \bf{answer}}$

$\bf{given} \div$

Time taken = 12 minutes.

Let the speed of car be x m/ minutes and AB = h , BC = y.

In ∆ABC,

$= > \: \frac{h}{y} = tan {45}^{0}$

$= > \frac{h}{y} = 1$

$= > h = y$

Now, in ∆ABD,

$= > tan {30}^{0} = \frac{h}{y + 12x}$

$= > \: \frac{1}{ \sqrt{3} } = \frac{h}{y + 12x}$

$= > \: \sqrt{3} h = y + 12x$

$=> \sqrt{3} y - y = 12x \: \: \: \: \: \: \: \: \: \: (h = y)$

$= > y = \frac{12x}{ \sqrt{3} - 1}$

$= > y = \frac{12x( \sqrt{3} + 1) }{2} \: \: \: \: \:$

$= > y = 6x( \sqrt{3} + 1)$

$\therefore \: time \: taken \: = 6x( \sqrt{3} + 1)min$

Answer 2

Step-by-step explanation:

refer to the attachment