On 19 June your vessel's 0523 ZT DR position is LAT 25°12.0'N, LONG 123°14.0'W, when an amplitude of the Sun is observed. The Sun's center is on the visible horizon and bears 052.0° per standard compass. Variation in the area is 15°E. The chronometer reads 01h 21m 58s and is 01m 18s slow. What is the deviation of the standard compass?
Official USCG Answer & Rule Citation
Correct Answer: Option D — 3.3°W
Celestial amplitude reduction yields the Sun's true bearing at visible sunset. Factoring chronometer time, position, and declination to find true azimuth, then applying 15°E variation against the standard compass bearing (052.0°psc), results in a compass deviation of 3.3°W.
100% Free & Open Access — No daily limits, paywalls, or gated answers on BrightMariner.