Question

As observed from the top of a lighthouse, 100 m above 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. [Take $\sqrt{3}=\mathrm{1.732.}$]

Answer 1

Given, AB= 100 m

Let x be the distance travelled by the ship during the period of observation. i.e., CD = x m

$In△ABDtan{60}^o=\frac{AB}{BD}\Rightarrow \sqrt{3}=\frac{100}{BD}\Rightarrow BD=\frac{100}{\sqrt{3}}-----\left(1\right)In△ABCtan{30}^o=\frac{AB}{BC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BD+CD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BD+x}\Rightarrow BD+x=100\sqrt{3}----\left(2\right)$

Put (1) in (2),

$\frac{100}{\sqrt{3}}+x=100\sqrt{3}x=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}=115.47m$ [Take $\sqrt{3}=1.732$ ]

Hence, the distance travelled by the ship during the period of observation is 115.47 m