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. 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, S_sa_² + _d_² - _b_² = 0; and if the equation of the supplements be (Σσα² + δ²) - β² = 0; then, because α² = 1 - _b_² = 1 - (S_sa_² + _d_²), and β² = 1 - _a_², substituting these values in the second equation, we get THEOREM III. (1 - Σσ × (1 - _d_²) - δ²) ⁄ (1 - S_s_Σσ) = _a_²; in words thus: