Answer to Question #101340 in Other for vERO

Question #101340
calculate the noon sun angle for the Eureka, California ON January 10. Use the terms darkness or tangent if applicable.
1
Expert's answer
2020-01-20T08:49:44-0500

Longitude of the city - "40^o 48'"

Accept the change in declination per day as "6'"

For 19 days the declination was changed on "19\\cdot 6=114'=1^o 54'" from winter solstice.

The height of the sun above the horizon on the day of the winter solstice


"90^o - 40^o 48' - 23^o 27'=25^o45'"

The noon sun angle at 10 January


"25^o45'+1^o54'=27^o39'"

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