Question

As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from ${30}^{\circ }$ to ${60}^{\circ }$. Find the distance travelled by the ship during the period of observation. (Use $\sqrt{3}=1.73$)

Answer 1

Let CD be a light house of length 100 m and A and B be the positions of ship sailing towards it.



Then, in $\Delta CBD$,

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

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

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

And, in $\Delta CAD$

$tan{30}^{\circ }=\frac{CD}{AC}$

$\frac{1}{\sqrt{3}}=\frac{100}{AC}$

$AC=100\sqrt{3}$

$\therefore$ Distance travelled by the ship $\left(AB\right)=AC-BC$

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

= $100\sqrt{3}\left(\frac{3-1}{3}\right)$

= $\frac{200\times 1.73}{3}$

= $115.33$ m