On 23 June in DR position LAT 21°39.0'S, LONG 106°28.0'W, you observe an amplitude of the Sun. The Sun's center is on the celestial horizon and bears 078°psc. The chronometer reads 02h 14m 39s and is 01m 43s slow. Variation in the area is 9°W. What is the deviation of the standard magnetic compass?

Correct Answer: Option A — 4.3°W

Calculate the Sun's amplitude using sin(Amp) = sin(Dec) / cos(Lat). Convert the amplitude to a true azimuth (Zn). Apply the variation to the true azimuth to find the magnetic bearing. The difference between the magnetic bearing and the compass bearing (078°psc) yields the deviation of 4.3°W, following standard celestial navigation procedures.