As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Step-by-Step Solution
Key Concept: Use the relation $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$ in the right‑angled triangles formed by the lighthouse, its base and each ship. The angle of depression from the top equals the angle of elevation from the ship, so $\tan\theta = \dfrac{\text{height of lighthouse}}{\text{horizontal distance}}$.
1. Draw a diagram: Let $L$ be the top of the lighthouse, $B$ its base on sea level, and $S_1$, $S_2$ the positions of the two ships such that $S_2$ is farther from the lighthouse than $S_1$ and both lie on the same straight line from $B$.
2. Identify right triangles: $\triangle LBS_1$ and $\triangle LBS_2$ are right‑angled at $B$.
3. Use angle of elevation: The angle of depression $\theta$ from $L$ to a ship equals the angle of elevation from the ship to $L$. Hence,
$$\tan\theta = \frac{\text{height of lighthouse}}{\text{horizontal distance from B to the ship}}.$$
4. Compute distances:
- For the ship with angle of depression $45^{\circ}$ (nearest ship):
$$d_1 = \frac{75}{\tan45^{\circ}} = \frac{75}{1} = 75\text{ m}.$$
- For the ship with angle of depression $30^{\circ}$ (farther ship):
$$d_2 = \frac{75}{\tan30^{\circ}} = \frac{75}{\frac{1}{\sqrt3}} = 75\sqrt3\text{ m}.$$
5. Find the distance between the ships:
$$\text{Distance} = d_2 - d_1 = 75\sqrt3 - 75 = 75(\sqrt3 - 1)\text{ m}.$$
6. Numerical value (optional):
$$75(\sqrt3 - 1) \approx 75(1.732 - 1) \approx 75(0.732) \approx 54.9\text{ m}.$$
Thus the two ships are about $55\text{ m}$ apart.
Correct Answer: $75(\sqrt{3} - 1)\text{ m} \;\approx\; 55\text{ m}$