= 0.50000, from which we get either A = 30.000° or (180° - 30.000°) = 150.000°.

In the present case, we know that the star is near the southern horizon, because the star is south of the equator and more than 2 hours east of the celestial meridian. Therefore, the second answer is correct, A = 150°. Many cases are less easy to decide by inspection. Equation (2.1) can be solved for cos A, and the correct quadrant then be determined. This is left to the student as a recommended exercise to gain experience in spherical astronomy! In the chapters to come (especially Chapter 6), we will make frequent use of the transformation relations to explore the possibilities of deliberate astronomical alignments of monuments.

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