Question

calculate the noon sun angle for the Eureka, California ON January 10. Use the terms darkness or tangent if applicable.

Answer 1

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

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