Question

From the top of the 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 degree in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time will the ship reach the light house.

Answer 1

From the top of the 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 degree in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time will the ship reach the light house.

the angle of depression 2

Let the top of the lighthouse is located at height H m.

When the angle of depression was $30{}^{\circ}$ , while the ship was at a distance from the lighthouse

$$S _ {1} = H \cot 3 0 {}^ {\circ} = H \sqrt {3}$$

When the angle of depression was $45{}^{\circ}$ , while the ship was at a distance from the lighthouse

$$S _ {2} = H \cot 4 5 {}^ {\circ} = H.$$

When the angle of depression was $90{}^{\circ}$ , while the ship reach the light house

$$S _ {3} = H \cot 9 0 {}^ {\circ} = 0.$$

Hence the distance changes from $H\sqrt{3}$ m to $H$ m in 10 minutes.

So the speed of the ship is

$$V = \frac {H \sqrt {3} - H}{1 0} \mathrm {m / m i n}$$

If a ship should reach the lighthouse, the distance varies from H m to 0.

Therefore

$$t = \frac {H - 0}{V} = \frac {1 0 H}{H \sqrt {3} - H} = \frac {1 0}{\sqrt {3} - 1} = 1 3. 6 6 \mathrm {m i n}$$

Answer: 13.66 min