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 QC_q_,
RC_r_. 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 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._