Question

A straight highway leads to the foot of a tower of a height $50$ m. From the top of the tower, the angles of depression of car standing on the highway are ${30}^0$ and ${60}^0$ respectively. What is the distance between the two cars and how far is each car from the tower?

Answer 1

Let $AB$ be the tower of height $50m$.

Let $C$ and $D$ be the position of $2$ cars.

Let $\angle ADB={60}^o$ and $\angle ACB={30}^o$

$\Rightarrow$ In $△ABD$, $\tan {60}^o=\frac{AB}{BD}$

$\Rightarrow$ $\sqrt{3}=\frac{50}{BD}$

$\therefore$ $BD=\frac{50}{\sqrt{3}}=\frac{50\sqrt{3}}{3}=\frac{50\times 1.732}{3}=28.86m$ $(Take\sqrt{3}=1.73)$

$\Rightarrow$ In $△ABC,\tan {30}^o=\frac{AB}{BC}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{50}{BC}$

$\Rightarrow$ $BC=50\sqrt{3}=86.6m$ $(Take\sqrt{3}=1.73)$

So, the distance of the first car from the tower $=28.86$

And the distance of the second car from the tower $=86.6m$

$\Rightarrow$ Now, distance between two cars $CD=BC-BD=86.6-28.86=57.74m$