On 11 January your vessel's 0655 zone time DR position is LAT 24°30'N, LONG 122°02'W, when an amplitude of the Sun is observed. The Sun's center is on the celestial horizon and bears 101.0° per standard compass. Variation in the area is 11.6°E. The chronometer reads 02h 52m 48s and is 02m 12s slow. What is the deviation of the standard compass?
Official USCG Answer & Rule Citation
Correct Answer: Option D — 1.4°E
Calculate the Sun's declination and LHA to determine the true amplitude. Compare the true amplitude to the compass bearing (101.0°psc) to find the total compass error. Subtract the 11.6°E variation from the compass error. The result is the deviation of the standard magnetic compass, which is 1.4°E.
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