Question

A man on top of a building observes a car at an angle of depression $\alpha$, where $tan\alpha =\frac{1}{\sqrt{5}}$ and sees that it is moving towards the base of the building. Ten minutes later the angle of depression of the car is found to be $\beta$ where tan $\beta =\sqrt{5}$. If the car is moving with an uniform speed, then how much more time will it take to reach the base of the tower?

Answer 1

Let 'h' be the height of the tower.
Let D be the point where the angle of depression from the top of the tower is ${\alpha }^{\circ }$ and C be the point where the angle of depression is $\beta$

$tan\alpha =\frac{1}{\sqrt{5}},tan\beta =\sqrt{5}$ ...... (given)

In $\Delta ABC$

$tan\beta =\frac{h}{x}$

$\sqrt{5}=\frac{h}{x}$

$x=\frac{h}{\sqrt{5}}\dots \left(i\right)$

In $\Delta ABD$

$tan\alpha =\frac{h}{y}$

$\frac{1}{\sqrt{5}}=\frac{h}{y}$

$y=h\sqrt{5}\dots \left(ii\right)$

To travel (y-x) i.e $(h\sqrt{5}-\frac{h}{\sqrt{5}})$ takes 10 minutes

$\therefore$ To travel $\frac{h}{\sqrt{5}}$ will take

$\frac{10\left(\frac{h}{\sqrt{5}}\right)}{h\sqrt{5}-\frac{h}{\sqrt{5}}}$

$=\frac{\frac{10h}{\sqrt{5}}}{\frac{5h-h}{\sqrt{5}}}$

$=\frac{10h}{4h}=2\frac{1}{2}$ minutes