Height & Distance — two boats due east with different angles of depression: From the top of a lighthouse, the angles of depression to two boats lying due east are 45° (nearer boat) and 30° (farther boat). The boats are 60 m apart along the same straight line. What is the height of the lighthouse (in meters)?

Introduction / Context:
Problems on angles of depression/elevation reduce to right triangles where tan(angle) = height / horizontal distance. With two sightlines from the same point, we can relate two horizontal distances to the same unknown height and then use their given separation to solve for height.

Given Data / Assumptions:

• Angle to nearer boat = 45°; angle to farther boat = 30°.
• Horizontal separation between boats = 60 m, collinear and due east of the lighthouse base.
• Let the lighthouse height be h meters; all ground is level.
  

Concept / Approach:
Let d₁ be horizontal distance to the 45° boat, and d₂ to the 30° boat. Using tan θ = h / d, we have d₁ = h / tan45° = h and d₂ = h / tan30° = h√3. The boats lie on the same ray, so d₂ − d₁ = 60.

Step-by-Step Solution:
d₂ − d₁ = h√3 − h = 60h(√3 − 1) = 60h = 60 / (√3 − 1) = 60(√3 + 1) / (3 − 1) = 30(√3 + 1)

Verification / Alternative check:
Numerically, √3 ≈ 1.732 ⇒ h ≈ 30 × 2.732 ≈ 81.96 m. Then d₁ ≈ 81.96 m and d₂ ≈ 141.96 m, whose difference is ≈ 60 m, matching the condition.

Why Other Options Are Wrong:
60√3 ≈ 103.92 m and 30(√3 − 1) ≈ 21.96 m do not satisfy d₂ − d₁ = 60 for consistent tangents. “90” is also inconsistent with the trigonometric relations.

Common Pitfalls:
Using d₁ + d₂ instead of the difference; confusing tan with cot; forgetting that angle of depression equals the angle of elevation at the ground point.

Final Answer:
30 ( √3 + 1 )