On 7 April in DR position LAT 27°42.0'N, LONG 114°03.0'W, you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 076°psc. The chronometer reads 02h 10m 17s and is 01m 52s slow. Variation in the area is 8°E. What is the deviation of the standard magnetic compass?
Official USCG Answer & Rule Citation
Correct Answer: Option D — 1.8°W
Calculate the Sun's amplitude using sin(Amp) = sin(Dec) / cos(Lat). With a True amplitude of 082.2° and a variation of 8°E, the compass error is 5.8°E. Subtracting the variation from the compass error (5.8°E - 8°E) results in a deviation of 1.8°W.
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