On 28 November your vessel's 0652 DR position is LAT 37°30'N, LONG 124°12'W, when an amplitude of the Sun is observed. The Sun's center is on the visible horizon and bears 103° per standard magnetic compass. Variation in the area is 16.3°E. The chronometer reads 02h 54m 18s and is 02m 06s fast. What is the deviation of the compass?

Correct Answer: Option A — 2.5°W

Calculate Greenwich Hour Angle and declination for the given date, then solve the amplitude equation $\sin(\text{Amp}) = \frac{\text{sin(Dec)}}{\cos(\text{Lat})}$ to find true sunrise/sunset bearing. Comparing true bearing against standard magnetic bearing and applying 16.3°E variation results in 2.5°W deviation.