A Navy captain in a boat is moving away from the foot of a vertical lighthouse at a constant speed of 4(√3 − 1) m/s. He observes that it takes 1 minute for the angle of elevation of the top of the lighthouse to change from 60° to 45°. What is the height (in metres) of the lighthouse?

Introduction / Context:
This problem combines motion in a straight line with trigonometric height and distance concepts. A Navy captain in a boat moves directly away from a lighthouse at a known constant speed. As he moves, the angle of elevation of the top of the lighthouse changes from 60° to 45° in a given time interval. Because the height of the lighthouse does not change, we can express the captain's varying horizontal distances from the lighthouse in terms of the height and the observed angles. The change in horizontal distance over the known time must match the distance travelled using speed multiplied by time, which allows us to solve for the unknown height of the lighthouse.

Given Data / Assumptions:

- The lighthouse is vertical and stands on level sea or ground. - The captain moves in a straight line directly away from the lighthouse. - Speed of the boat is 4(√3 − 1) m/s. - Time taken for the angle of elevation to change from 60° to 45° is 1 minute (60 seconds). - At the initial position, angle of elevation of the top of the lighthouse is 60°. - At the later position, angle of elevation is 45°. - We use tan 60° = √3 and tan 45° = 1.

Concept / Approach:
Let H be the height of the lighthouse. At the first observation point, the captain is at a horizontal distance x from the foot of the lighthouse, forming a right angled triangle with height H. At the second observation point, after moving further away, he is at a distance x2 from the foot. From the definition of tangent, tan 60° = H / x and tan 45° = H / x2. These give x = H / √3 and x2 = H. The captain travels from x to x2 in 60 seconds, so the increase in distance x2 − x must equal speed multiplied by time, that is 4(√3 − 1) * 60. Equating these leads to a simple equation in H that we can solve.

Step-by-Step Solution:
Step 1: Let H be the height of the lighthouse in metres. Step 2: At the first position, angle of elevation is 60°, so tan 60° = H / x, where x is the initial horizontal distance. Step 3: Since tan 60° = √3, we get √3 = H / x, so x = H / √3. Step 4: At the second position, the angle of elevation is 45°, so tan 45° = H / x2, where x2 is the new horizontal distance. Step 5: Because tan 45° = 1, we get 1 = H / x2, so x2 = H. Step 6: The captain travels from distance x to distance x2 in 60 seconds, so the increase in horizontal distance is x2 − x = H − H / √3. Step 7: His speed is 4(√3 − 1) m/s, so the distance travelled in 60 seconds is 4(√3 − 1) * 60 = 240(√3 − 1). Step 8: Equate the two expressions for the distance travelled: H − H / √3 = 240(√3 − 1). Step 9: Factor H on the left: H(1 − 1 / √3) = 240(√3 − 1). Step 10: Write 1 − 1 / √3 as (√3 − 1) / √3. Then H * (√3 − 1) / √3 = 240(√3 − 1). Step 11: Cancel the common factor (√3 − 1) from both sides, giving H / √3 = 240. Step 12: Multiply both sides by √3 to get H = 240√3. Step 13: Thus, the height of the lighthouse is 240√3 m.

Verification / Alternative check:
We can verify the result numerically. Using √3 ≈ 1.732, H ≈ 240 * 1.732 ≈ 415.7 m. Then x = H / √3 ≈ 415.7 / 1.732 ≈ 240 m and x2 = H ≈ 415.7 m. The distance travelled is x2 − x ≈ 415.7 − 240 ≈ 175.7 m. Now compute the distance using speed and time: speed = 4(√3 − 1) ≈ 4(1.732 − 1) ≈ 4 * 0.732 ≈ 2.928 m/s. In 60 seconds, distance ≈ 2.928 * 60 ≈ 175.7 m, which matches the geometry, confirming that H = 240√3 m is consistent.

Why Other Options Are Wrong:
Option B (480(√3 − 1)) represents the total distance travelled, not the height of the lighthouse. Option C (360√3) and option D (280√2) do not satisfy the relationship between the two tangent equations and the distance travelled when substituted into the same reasoning. Only 240√3 m simultaneously satisfies both the trigonometric conditions and the motion constraint given by the speed and time.

Common Pitfalls:
Students often confuse the roles of x and x2, sometimes treating both as equal or forgetting that the second distance must be greater. Another common error is to use the wrong trigonometric ratio (such as sine or cosine) instead of tangent when relating height to horizontal distance. Algebraic mistakes can also occur when simplifying 1 − 1 / √3. Carefully expressing the problem in terms of tan θ = height / base at each point and explicitly equating the change in distance to speed multiplied by time helps prevent these issues.

Final Answer:
The height of the lighthouse is 240√3 m, which corresponds to option A.