Question

From the top of a tower 100 m high, a man observes two cars on the opposite of the tower with angles of depression ${30}^{\circ }and{45}^{\circ }$ respectively. Find the distance between the cars.
$Use[\sqrt{3}=1.73]$

Answer 1

The given information can be diagrammatically represented as,

Here, AB is the tower of height 100 m. The Points C and D are the position of the two cars.
In right $\Delta$ACB,
$tan{45}^{\circ }=\frac{AB}{BC}$
$\Rightarrow 1=\frac{100m}{BC}$
$\Rightarrow BC=100m$
$Inright\Delta ABD$
$tan{30}^{\circ }=\frac{AB}{BD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100m}{BD}$
$\Rightarrow BD=100\sqrt{3}m$
Distance between the two cars = CD
= BC + CD
$=100m+100\sqrt{3}m$
$=100m+100\times 1.73m$
=100m+173m
= 273m
Thus, the distance between two cars is 273 m.