From the top of a light house, it is observed that a ship is sailing directly towards it and the angle of depression of the ship changes from 30° to 45° in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time (in minutes) will the ship reach the light house ?

Let AB be the lighthouse of height h meters and C and D be the position of the ship.

From figure,

∠ADC = ∠EAD = 30° (Alternate angles are equal)

∠ACB = ∠EAC = 45° (Alternate angles are equal)

From figure,

In right angle triangle ADB

⇒ tan 30° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD}$

⇒ BD = $\sqrt{3}\text{AB} = \sqrt{3}h$.

In right angle triangle ABC

⇒ tan 45° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

⇒ 1 = $\dfrac{AB}{BC}$

⇒ BC = AB = h.

From figure,

CD = BD - BC = $\sqrt{3}h - h = h(\sqrt{3} - 1)$ meters.

Given,

Angle of depression of the ship changes from 30° to 45° in 10 minutes.

∴ Ship comes from position D to C in 10 minutes.

∴ Ship covers a distance of CD = $h(\sqrt{3} - 1)$ in 10 minutes.

Speed of ship = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{h(\sqrt{3} - 1)}{10}$ meter/minute.

Time taken by ship to cover (BC = h meters) is

$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}} \\[1em] = \dfrac{h}{\dfrac{h(\sqrt{3} - 1)}{10}} \\[1em] = \dfrac{10}{\sqrt{3} - 1} \\[1em] = \dfrac{10}{1.732 - 1} \\[1em] = \dfrac{10}{0.732} \\[1em] = 13.66 \text{ minutes}.$

Hence, ship will reach the lighthouse in 13.66 minutes.