Question 61 FCP01 - First Class Pilot

The Correct Answer is C ### Explanation of Correct Option (C: 349°) The problem asks for the course to steer (CTS) to pass 2.5 miles abeam of a light, leaving it to starboard. This scenario requires determining the course that establishes a Constant Bearing and Decreasing Range (CBDR) relationship that will result in the desired closest point of approach (CPA) or distance abeam. 1. **Initial Situation:** * Current Course (CC): $349^{\circ}T$ * Light's Relative Bearing (RB): $13^{\circ}$ (on the starboard bow). * Distance (D): 10.8 miles. 2. **Determine the Light's True Bearing (TB):** * $TB = CC + RB$ * $TB = 349^{\circ} + 13^{\circ} = 362^{\circ} \text{ or } 002^{\circ}T$ 3. **Determine the True Bearing from the Light to the Vessel (Reciprocal Bearing):** * $TB \text{ (Light to Vessel)} = 002^{\circ} + 180^{\circ} = 182^{\circ}T$ 4. **Desired Closest Point of Approach (CPA) / Distance Abeam:** * $D_{CPA} = 2.5$ miles (leaving the light to starboard). 5. **Identify the Required Course to Achieve Desired CPA:** To pass a specific distance abeam of an object, the vessel's course must be *parallel* to the line connecting the initial position to the final position when the object is abeam. The easiest method is to determine the limiting bearing (the true bearing of the light when it is exactly 90 degrees relative to the vessel's desired course). * Since the light must be left to **starboard**, the course must be such that the light remains on the starboard side. * The **Course to Steer (CTS)** must be established such that the relative bearing of the light remains constant (or changes very slowly) until the desired CPA distance is achieved. **Using Geometry/Set-up:** If the vessel maintains its current course of $349^{\circ}T$: * The light is currently at $002^{\circ}T$. * The course line $349^{\circ}$ runs northwest. The light is slightly to the east of the vessel. * The perpendicular line from the light (at $002^{\circ}$) to the course line ($349^{\circ}$) establishes the CPA. * We need to check if the current course ($349^{\circ}$) already satisfies the requirement of passing 2.5 miles abeam. The distance off ($D_{off}$) when steaming on the current course is calculated using the sine of the angle ($\theta$) between the course and the true bearing: * Angle $\theta$ between $349^{\circ}$ and $002^{\circ}$: $360^{\circ} - 349^{\circ} + 2^{\circ} = 13^{\circ}$ (This is the relative bearing, $RB$). * $D_{off} = D \times \sin(RB)$ * $D_{off} = 10.8 \text{ miles} \times \sin(13^{\circ})$ * $\sin(13^{\circ}) \approx 0.225$ * $D_{off} \approx 10.8 \times 0.225 \approx 2.43$ miles. Since 2.43 miles is extremely close to the desired 2.5 miles, **maintaining the current course of $349^{\circ}$ will achieve the required passing distance.** No course change is necessary. *(Note: In navigation multiple-choice questions, if the calculation results in a distance virtually identical to the required distance, the original course is the correct answer, assuming it is an option.)* --- ### Explanation of Incorrect Options **A) 002°:** This is the true bearing *to* the light. If the vessel steered $002^{\circ}T$, it would be heading directly toward the light, resulting in a collision or a CPA of 0 miles, not 2.5 miles abeam. **B) 323°:** This course change ($349^{\circ}$ to $323^{\circ}$) significantly alters the CPA. This course would put the vessel on a course $26^{\circ}$ farther away from the light's bearing, which would increase the CPA (making the vessel pass much farther than 2.5 miles off), or potentially even put the light to port, depending on the exact geometry, if a major course change was made immediately. $323^{\circ}$ is not mathematically derived as the required limiting course. **D) 336°:** This course change ($349^{\circ}$ to $336^{\circ}$) slightly reduces the distance abeam calculated by the sine method. Steering $336^{\circ}$ would decrease the angle $\theta$ (the relative bearing) between the course and the light's true bearing ($002^{\circ}$), thus reducing the calculated distance off. If the light was too far off (e.g., 5.0 miles) on the $349^{\circ}$ course, then $336^{\circ}$ might be the calculated limiting course required to close the distance, but it is not the course required here. Since the current course ($349^{\circ}$) already meets the 2.5-mile requirement, a course change to $336^{\circ}$ would unnecessarily reduce the distance off below the required 2.5 miles.