Angle of depression + speed → time to reach directly beneath From a bridge 15 m above a river, the angle of depression to a boat is 30°. If the boat moves at 6 km/h on a straight path toward the bridge's vertical, how long (in seconds) will it take to be exactly beneath the bridge?

Introduction / Context:Angle-of-depression scenarios translate to angle-of-elevation from the boat to the bridge top. The right triangle’s vertical leg is fixed (bridge height), which gives the horizontal distance via tan. With the boat’s uniform speed, time = distance/speed after unit consistency.

Given Data / Assumptions:

• Bridge height h = 15 m (vertical).
• Angle of depression = 30° ⇒ tan 30° = h/d.
• Boat speed = 6 km/h (straight toward the point below the bridge).

Concept / Approach:Horizontal distance d = h / tan 30° = 15 / (1/√3) = 15√3 m. Convert 6 km/h to m/s, then compute time t = d / v.

Step-by-Step Solution:

Verification / Alternative check:Compute numerically: √3 ≈ 1.732 ⇒ 9√3 ≈ 15.588 s. Distance 15√3 ≈ 25.98 m at 1.666… m/s takes ≈ 15.59 s, consistent.

Why Other Options Are Wrong:19/√3 s or 3√3 s do not satisfy d = v * t with the given geometry and speed.

Common Pitfalls:Using sin or cos instead of tan for horizontal distance, or failing to convert km/h to m/s before computing time.

Final Answer:9 √3 second