B3 - - THE OBLIQUE SPHERICAL TRIANGLE - - 10/03/2004

LAW OF SINES
sin A / sin a = sin B / sin b = sin C / sin c

LAW OF COSINES
cos a = cos b * cos c + sin b * sin c * cos A

cos A = -cos B * cos C + sin B * sin C * cos a

NAPIERS ANALOGIES
tan ((A-B)/2) = ctn (C/2) * sin ((a-b)/2) / sin ((a+b)/2)

tan ((A+B)/2) = ctn (C/2) * cos ((a-b)/2) / cos ((a+b)/2)

            A = (( A + B)/2 ) + (( A - B )/2)

            B = (( A + B)/2 ) - (( A - B )/2)

tan ((a-b)/2) = tan (c/2) * sin ((A-B)/2) / sin ((A+B)/2)

tan ((a+b)/2) = tan (c/2) * cos ((A-B)/2) / cos ((A+B)/2)

The final answer to a spherical trigonometry problem must give a value between 0 and 180 degrees (first or second quadrant).

When the final answer comes from the tangent function it may be a negative angle (fourth quadrant). If so, add 180 degrees to the angle to put the result into the second quadrant.

When the answer comes from the sine function the answer is ambiguous. The answer could be in the first or in the second quadrant.

In addition to the above equations, it is sometimes possible to drop a perpendicular from one corner of the triangle to the opposite side, creating two right triangles. This will result in more equations to solve, but the calculations will be simplified. In the next section there are two examples of this method, the Ageton method (page C2), and the no-name method (page C4).

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