Two angles of depression to ships 200 m apart → lighthouse height From the top of a lighthouse, the angles of depression of two ships on the same straight line eastward are 45° and 30°. If the ships are 200 m apart along that line, find the height of the lighthouse.

Introduction / Context:
Angles of depression equal angles of elevation from the horizontal at the ship. With both ships along the same ground line from the lighthouse foot (east), their horizontal distances differ by 200 m. Using tan relations for 45° and 30°, we solve for the common height.

Given Data / Assumptions:

• Angles of depression: nearer ship 45°, farther ship 30° (or vice versa—but distances will sort themselves since 45° corresponds to the nearer one).
• Distance between ships along the line = 200 m.
• Let height be h and horizontal distances be d₁ (for 45°), d₂ (for 30°), with d₂ > d₁.

Concept / Approach:
tan 45° = h/d₁ ⇒ d₁ = h. tan 30° = h/d₂ ⇒ d₂ = h√3. The ships are 200 m apart, so d₂ − d₁ = 200 ⇒ h√3 − h = 200.

Step-by-Step Solution:

Verification / Alternative check:
Distances: d₁ ≈ 273.2 m; d₂ ≈ 473.2 m; difference ≈ 200 m, consistent.

Why Other Options Are Wrong:
100 m and 173 m understate the geometry; 200 m is the separation, not the height.

Common Pitfalls:
Assigning the 30° to the nearer ship (which would invert distances), or forgetting to rationalize when comparing to rounded options.

Final Answer:
273 m