Let QR (Tab. [XX.] Fig. 1.) be the base of a spherical triangle; its sides PQ, PR, whose planes cut that of the base in the diameters QCq, RCr. And if, from the angle P, the line PL is perpendicular to the plane of the base, meeting it in L, all planes drawn through PL will be perpendicular to the same, by 18. el. 11. Let two such planes be perpendicular likewise to the semicircles of the sides, cutting them in the straight lines PG, PH; and the plane of the base in the lines LG, LH.
Then the plane of the triangle PGL being perpendicular to the two planes, whose intersection is QGCq, the angles PGQ LGQ will be right angles, by 19. el. 11. PG likewise subtends a right angle PLG, and the angle PGL measures the inclination of the semicircle QPq to the plane of the base (def. 6. el. 11.) that is (by 16 el. 3. and 10 el. 11.) it is equal to the spherical angle PQR: whence PG is to PL as the radius to the sine of PQR. The same way PL is to PH as the sine of PRQ is to the radius: and therefore, ex æquo. PG the sine of the side PQ is to PH the sine of PR, as the sine of PRQ is to the sine of PQR.
CASES II. and III.
When the three parts are of the same name.
And,
When two given parts include between them a given part of a different name, the part required standing opposite to this middle part.
Theorem II.
Let S and s be the sines of two sides of a spherical triangle, d the sine of half the difference of the same sides, a the sine of half the included angle, b the sine of half the base; and writing unity for the radius, we have Ssa² + d² - b² = 0; in which a or b may be made the unknown quantity, as the case requires.
Demonstration.
Let PQR ([Fig. 2.]) be a spherical triangle, whose sides are PQ PR, the angle included QPR, the base QR, PC the semiaxis of the sphere, in which the planes of the sides intersect.