Then the plane of the triangle PGL being perpendicular to the two planes, whose intersection is QGC_q_, 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 QP_q_ 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. To the pole P, draw the great circle AB, cutting the sides (produced, if needful) in M and N; and thro’ Q and R, the lesser circles Q_q_, _r_R, cutting off the arcs Q_r_ _q_R equal to the difference of the sides; join MN, Q_q_, _r_R, QR, _qr_.