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_.
Then the planes of the circles described being parallel (_Theod.
sphæric._ 2. 2.), and the axis PC perpendicular to them (_10. 1. of
the same_), their intersections with the planes of the sides, as QT,
and R_t_, will make right angles with PC; that is, QT and R_t_ are the
sines (S, _s_.) of the sides PQ PR, and MC NC are whole sines. Now the
isosceles triangles MCN, QT_q_, _rt_R, being manifestly similar; as
also MN, the subtense of the arc which measures the angle QPR, being
equal to (2_a_) twice the sine of half that angle; we shall have MN:
MC∷ Q_q_: QT∷ _r_R: R_t_; or, in the notation of the theorem, Q_q_ =
2S_a_, _r_R = 2_sa_. And further, the chords Q_r_ _q_R being equal, and
equally distant from the center of the sphere, as also equally inclined
to the axis PC, will, if produced, meet the axis produced, in one point
Z. Whence the points Q, _q_, R, _r_, are in one plane (2. _el._ 11.),
and in the circumference in which that plane cuts the surface of the
sphere: the quadrilateral Q_q_R_r_ is also a segment of the isosceles
triangle ZQ_q_, cut off by a line parallel to its base, making the
diagonals QR, _qr_, equal. And therefore, by a known property of the
circle, Q_q_ × _r_R + (_q_R)² = (QR)²; which, substituting for Q_q_ and
R_r_ the values found above, 2_d_ for Q_r_, 2_b_ for QR, and taking the
fourth part of the whole, becomes S_sa_² + _d_² = _b_² the proposition
that was to be demonstrated.
_Note_ 1. If this, or the preceding, is applied to a plane triangle,
the sines of the sides become the sides themselves; the triangle
being conceived to lie in the surface of a sphere greater than any
that can be assigned.
_Note_ 2. If the two sides are equal, _d_ vanishing, the operation
is shorter: as it likewise is when one or both sides are quadrants.
_Note_ 3. By comparing this proposition with that of the Lord
Neper[26], which makes the 39th of Keill’s Trigonometry, it appears,
that if AC, AM, are two arcs, then sin. (AC + AM) ⁄ 2 × sin. (AC -
AM) ⁄ 2 = ((_b_ + _d_) × (_b_ - _d_) =) (sin. ½ AC + sin. ½ AM) ×
(sin. ½ AC - sin. ½ AM). And in the solution of Case II. the first of
these products will be the most readily computed.