Question 60 FCP01 - First Class Pilot

The Correct Answer is D ### 1. Explanation for Option D (222°T) This problem requires calculating the new course (Course 2) needed to achieve a desired CPA (Closest Point of Approach) or beam distance relative to the daymark, based on a change of course (alteration) executed at a specific time (1009). **Step 1: Determine the relative position of the daymark at the time of the course change (1009).** * **Initial Data (1006):** Course 1 = $241^{\circ} \text{T}$, Speed = $18.2$ knots, Range to Daymark = $3.9$ miles, Bearing to Daymark = $241^{\circ} \text{T}$. * Since the bearing and the course are identical ($241^{\circ} \text{T}$), the vessel is heading directly toward the daymark. * **Time interval ($\Delta T$):** $1009 - 1006 = 3$ minutes. * **Distance traveled ($D_{1}$):** $D_{1} = \text{Speed} \times \frac{\Delta T}{60}$ $D_{1} = 18.2 \text{ knots} \times \frac{3}{60} \text{ hours} = 18.2 \times 0.05 = 0.91$ miles. * **Range at 1009 ($R_{2}$):** The daymark remains directly ahead. $R_{2} = \text{Initial Range} - D_{1}$ $R_{2} = 3.9 \text{ miles} - 0.91 \text{ miles} = 2.99$ miles. At 1009, the vessel is $2.99$ miles from the daymark, bearing $241^{\circ} \text{T}$. **Step 2: Calculate the required course change angle ($\theta$).** The goal is to leave the daymark abeam (at the CPA) to starboard at a distance of $1.0$ mile. * The relationship between the range at alteration ($R_{2}$), the desired CPA ($D_{CPA}$), and the angle of turn off the initial line of approach ($\theta$) is given by the formula based on trigonometry: $$\sin(\theta) = \frac{D_{CPA}}{R_{2}}$$ * $D_{CPA} = 1.0$ mile * $R_{2} = 2.99$ miles $$\sin(\theta) = \frac{1.0}{2.99} \approx 0.3344$$ $$\theta = \arcsin(0.3344) \approx 19.54^{\circ}$$ **Step 3: Determine the New Course (Course 2).** The vessel needs to leave the daymark abeam to starboard. * The initial bearing to the daymark (the line of approach) is $241^{\circ} \text{T}$. * To leave the mark to starboard, the vessel must turn to port (left) relative to the daymark's initial position. This means the new course must be less than the initial course ($241^{\circ} \text{T}$). * New Course (Course 2) = Initial Bearing (Line of Approach) - $\theta$ Course 2 = $241^{\circ} \text{T} - 19.54^{\circ}$ Course 2 $\approx 221.46^{\circ} \text{T}$ Rounding to the nearest degree gives $221^{\circ} \text{T}$ or $222^{\circ} \text{T}$. Option D ($222^{\circ} \text{T}$) is the closest and correct answer. *** ### 2. Explanation of why other options are incorrect **A) 257°T:** This option corresponds to turning to starboard (right) by $16^{\circ}$. If the vessel turns to starboard, it will leave the daymark to port. $241^{\circ} \text{T} + 19.54^{\circ} \approx 260.54^{\circ} \text{T}$. $257^{\circ} \text{T}$ is too far off the required course (either $222^{\circ} \text{T}$ or $261^{\circ} \text{T}$) and is based on turning the wrong way (to port relative to the required starboard passage). **B) 218°T:** This course implies a port turn of $241^{\circ} - 218^{\circ} = 23^{\circ}$. While this is the correct direction (port turn to leave the mark to starboard), the calculated required angle ($\theta$) is $19.54^{\circ}$. A $23^{\circ}$ turn would result in a CPA of: $D_{CPA} = 2.99 \times \sin(23^{\circ}) \approx 2.99 \times 0.3907 \approx 1.17$ miles. This CPA is greater than the required $1.0$ mile, making it incorrect. **C) 260°T:** This course implies a starboard turn of $260^{\circ} - 241^{\circ} = 19^{\circ}$. This turn angle ($19^{\circ}$) is very close to the required angle ($\theta = 19.54^{\circ}$). However, turning to $260^{\circ} \text{T}$ means the vessel is turning to starboard, which will leave the daymark **to port**, not abeam to starboard as required by the question. The course corresponding to a turn of $19.54^{\circ}$ to starboard would be $261^{\circ} \text{T}$.