Question

As observer from the top of a lighthouse $100$ $\text{m}$ high observe sea level, the angle of depression of a ship sailing directly towards it, changes from ${30}^{\circ }$ to ${60}^{\circ }$ . Determine the distance travelled by the ship during the period of observation.

Answer 1

In $△ABD$,

$\therefore \tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BD}$

$\Rightarrow BD=100\sqrt{3}\text{m}$

In $△ABC$,

$\tan {60}^{\circ }=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{100}{BC}$

$\Rightarrow BC=\frac{100}{\sqrt{3}}$

$\therefore$ Distance travelled by the ship$=CD=BD-BC$

$=(100\sqrt{3}-\frac{100}{\sqrt{3}})$

$=100\left(\frac{2}{\sqrt{3}}\right)$

$=\frac{200}{\sqrt{3}}\text{m}$