Extend both meridians past fourth side D of quadrilateral ABCD, to north pole P --- area of resulting semi-lune equals angle P = side A. Consider triangle PQR remaining after subtracting ABCD from semi-lune, with sides D, pi/2-B, pi/2-C --- via cosine rule, cos D = sin B sin C + cos B cos C cos A ; via sine rule, sin Q = cos B sin A csc D ; sin R = cos C sin A csc D ; area PQR = A + Q + R - pi ; area ABCD = A - PQR = pi - Q - R . WFL On 2/7/11, David Wilson <davidwwilson@comcast.net> wrote:

On the unit sphere, measuring angles in radians.

Place an arc with measure A on the equator. Place arcs with measures B and C at each end of A running north along meridians (perpendicular to A). Join the northern endpoints of B and C with a great circle arc. What is the area of the spherical quadrilateral thus formed?

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