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Looking at the drawing above, the angular change desired is indicated by "delta question mark". Can you use trigonometry to determine the desired angular change? The answer for latitude 42 degrees north (Chicago) is about 31.6 degrees. This by the way is the angle that a shadow casts south of east, at sunset, for any tree or street pole on that day.
The Sun's declination changes +23.5 degrees from the Vernal Equinox day to the Summer Solstice Day, and this change is equal to the perpendicular angular change between the Sun's celestial paths on those two days. What is interesting is what the change in the azimuthal angle is between the points of impingement on those two days. In other words, how many degrees north of west does the Sun set on the Summer Solstice day for your latitude?
On an equinox day, such as the Vernal Equinox day, the point of impingement is known, because the Sun sets "cardinal west" everywhere on Earth on that day, so if you see the Sun a few minutes before sunset, you can form the angle shown to the left in your mind, and thereby determine your latitude.
Another way would be to determine the angle between the Sun's celestial path (its actual movement in the sky) at sunset, the point the Sun strikes on the westerly horizon (the point of impingement), and the vertical up from that point. The angle thus formed is your latitude.
One way would be to use the North Star and a sextant (glorified protractor) in the Northern Hemisphere, or the Southern Cross and a sextant in the Southern Hemisphere.
Suppose you are lost on a deserted island, with just a Fedex box and a volleyball. How would you determine your latitude?
At the Equator, of course, your latitude is 0 degrees, so the Sun should descend vertically, not only on an Equinox Day, but every day. The image on the left shows the Sun's celestial path for three days: the Winter Solstice day, the Vernal Equinox Day, and the Summer Solstice Day.
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