Philosophical transactions, Vol. L. Part II. For the year 1758. / Giving some account of the present undertakings, studies, and labours, of the ingenious, in many considerable parts of the world.

by Various; Royal Society

CASE I.

When of three given parts two stand opposite to each other, and the third stands opposite to the part required.

Theorem I.

The sines of the sides are proportional to the sines of angles opposite to them.

Demonstration.

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 QCq, RCr. 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 QGCq, 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 QPq 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 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.