Posted by **John - PLEASE Help!** on Monday, April 11, 2011 at 1:36am.

After shooting the trailing edge of the Sun on 11 April 2008, the observed clockwise angle from the mark to the edge of the sun is 259 degrees- 44 minutes- 51 seconds, the latitude is 32 degrees- 47 minutes- 54 seconds North, the longitude is 079 degrees-57minutes- 38 seconds, the time is 11 hours- 17 minutes - 30 seconds Daylight Savings time. The LaPlace correction is 3seconds. Determine the geodetic bearing to the mark.

- Geomatics -
**MathMate**, Monday, April 11, 2011 at 8:58am
Briefly, you will need access to the Nautical Almanac, published jointly by UK and US yearly. Some free versions are available on the web for past years.

The nautical almanac gives the position of the sun (and many other celestial bodies) in longitude and latitude at any hour of the year.

Most of the time, this, together with the time of the day, is used to calculate the position of the observer. In your case, the observer's position is known, so is that of the sun. So you can calculate the observed bearing of the sun using spherical trigonometry, and hence that of the mark.

## Answer this Question

## Related Questions

- physical science - The diameter of the sun makes an angle of .53 degrees from ...
- Physics - Astronomers find that light emitted by a particular element at one ...
- physics - the solar energy arriving at the outer edge of earth's atmosphere from...
- trig - At the winter solstice, the power, in watts, received from the Sun on ...
- PHYSICS - An underwater scuba diver sees the sun at an apparent angle of 45 ...
- Earth Science - Please determine the noon sun angle at the given latitudes on ...
- physics - How is the angle of inclination of the sun measured on the vernal or ...
- physics - I keep getting the wrong answer... A solar model is used to calculate ...
- College Chemistry - I am trying to figure out how many minutes are required for ...
- physics - A meter stick lies on the bottom of a 100 cm long tank with its zero ...