Question

A person on the top of a 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 ${30}^{\circ }to{45}^{\circ }$ how soon after this , will the car reach the tower?

Answer 1

Consider the triangle ABD

${\tan 30}^o=\frac{AB}{BD}$

$\frac{1}{\sqrt{3}}=\frac{AB}{BD}$

$\frac{BD}{\sqrt{3}}=AB$

$\frac{BC+CD}{\sqrt{3}}=AB⟹\left(1\right)$

consider the triangle ABC

${\tan 45}^o=\frac{AB}{BC}$

$1=\frac{AB}{BC}$

$AB=BC⟹\left(2\right)$

From equation 1 and 2

$\frac{AB+CD}{\sqrt{3}}=AB$

$\sqrt{3}AB-AB=CD$

$AB(\sqrt{3}-1)=CD⟹\left(3\right)$

Let v m/min be the speed of car

Since the car takes 12 mins to cover the distance CD,we have

$CD=12v⟹\left(4\right)$

From equation 4 and 3,we have

$AB(\sqrt{3}-1)=12v$

$\frac{AB(\sqrt{3}-1)}{12}=v⟹\left(5\right)$

$time=\frac{BC}{v}$

$time=\frac{12BC}{AB(\sqrt{3}-1)}and\frac{12AB}{AB(\sqrt{3}-1)}$

$time=\frac{12}{(\sqrt{3}-1)}=16.4mins$