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 Qq, rR, cutting off the arcs Qr qR equal to the difference of the sides; join MN, Qq, rR, 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 Rt, will make right angles with PC; that is, QT and Rt are the sines (S, s.) of the sides PQ PR, and MC NC are whole sines. Now the isosceles triangles MCN, QTq, rtR, being manifestly similar; as also MN, the subtense of the arc which measures the angle QPR, being equal to (2a) twice the sine of half that angle; we shall have MN : MC ∷ Qq : QT ∷ rR : Rt; or, in the notation of the theorem, Qq = 2Sa, rR = 2sa. And further, the chords Qr qR 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 QqRr is also a segment of the isosceles triangle ZQq, cut off by a line parallel to its base, making the diagonals QR, qr, equal. And therefore, by a known property of the circle, Qq × rR + (qR)² = (QR)²; which, substituting for Qq and Rr the values found above, 2d for Qr, 2b for QR, and taking the fourth part of the whole, becomes Ssa² + d² = b² 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.
CASE IV.
When the part required stands opposite to a part, which is likewise unknown: Having from the data of Case I. found a fourth part, let the sines of the given sides be S, s; those of the given angles Σ, σ; and the sines of half the unknown parts a and b; and we shall have, as before, Ssa² + d² - b² = 0; and if the equation of the supplements be Σσα² + δ² - β² = 0; then, because α² = 1 - b² = 1 - (Ssa² + d²), and β² = 1 - a², substituting these values in the second equation, we get
Theorem III.
1 - Σσ × (1 - d²) - δ² ⁄ 1 - SsΣσ = a²; in words thus:
Multiply the product of the sines of the two known angles by the square of the cosine of half the difference of the sides: add the square of the sine of half the difference of the angles; and divide the complement of this sum to unity, by the like complement of the product of the four sines of the sides and angles; and the square root of the quotient shall be the sine of half the unknown angle.