PROP. XV.

Ir the hypothenuse of a right angled triangle be greater or less than a quadrant, the sides will be of different or the same affection.

This is the converse of the preceding, and demonstrated in the

same manner.

PROP. XVI.

IN any spherical triangle ABC, if the perpendicular AD from A on the base BC, fall within the triangle, the angles B and C at the base will be of the same affection; and if the perpendicular fall without the triangle, the angles B and C will be of different affection.

Fig. 11.-1. Let AD fall within the triangle; then (13. of this), since ADB, ADC are right angled spherical triangles, the angles B, C, must each be of the same affection as AD.

Fig. 12.-2. Let AD fall without the triangle; then (13. of this), the angle B is of the same affection as AD; and by the same the angle ACD is of the same affection as AD; therefore the angles ACB and AD are of different affection, and the angles B and ACB of different affection.

COR. Hence if the angles B and C be of the same affection, the perpendicular will fall within the base; for, if it did not (16. of this), B and C would be of different affection. And if the angles B and C be of opposite affection, the perpendicular will fall without the triangle; for, if it did not (16. of this), the angles B and C would be of the same affection, contrary to the supposition.

PROP. XVII. FIG. 13.

IN right angled spherical triangles, the sine of either of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side.

Let ABC be a triangle, having the right angle at A; and let AB be either of the sides; the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC. Let D be the centre of the sphere; join AD, BD, CD, and let AE be drawn perpendicular to BD, which therefore will be the sine of the arch AB, and from the point E, let there be drawn in the plane BDC the straight line EF at right angles to BD, meeting DC in F, and let AF be joined. Since therefore the straight line DE is at right angles to both EA and EF, it will also be at right angles to the plane AEF (4. 11.); wherefore the plane ABD, which passes through DE, is perpendicular to the plane AEF (18. 11.) and the plane AEF perpendicular to ABD; the plane ACD or AFD is also perpendicular to the same ABD :

therefore the common section, viz. the straight line AF, is at right angles to the plane ABD (19. 11.), and FAE, FAD are right angles (3. def. 11.); therefore AF is the tangent of the arch AC; and in the rectilineal triangle AEF, having a right angle at A, AE will be to the radius as AF to the tangent of the angle AEF (1. Pl. Tr.); but AE is the sine of the arch AB, and AF the tangent of the arch AC, and the angle AEF is the inclination of the planes CBD, ABD (6. def. 11.), or the spherical angle ABC: therefore the sine of the arch AB is to the radius, as the tangent of the arch AC, to the tangent of the opposite angle ABC.

COR. 1. If therefore of the two sides, and an angle opposite to one of them, any two be given, the third will also be given.

COR. 2. And since by this proposition the sine of the side AB is to the radius, as the tangent of the other side AC to the tangent of the angle ABC opposite to that side; and as the radius is to the cotangent of the angle ABC, so is the tangent of the same angle ABC to the radius (Cor. 2. def. Pl. Tr.); by equality, the sine of the side AB is to the co-tangent of the angle ABC adjacent to it, as the tangent of the other side AC to the radius.

PROP. XVIII. FIG. 13.

In right angled spherical triangles, the sine of the hypothenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side.

Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypothenuse BC will be to the radius, as the sine of the arch AC is to the sine of the angle ABC.

Let D be the centre of the sphere, and let CG be drawn perpendicular to DB, which will therefore be the sine of the hypothenuse BC; and from the point G let there be drawn in the plane ABD the straight line GH perpendicular to DB, and let CH be joined: CH will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line FA; wherefore CHD, CHG are right angles, and CH is the sine of the arch AC; and in the triangle CHG, having the right angle CHG, CG is to the radius, as CH to the sine of the angle CGH (1. Pl. Tr.); but since CG, HG are at right angles to DBG, which is the common section of the planes CBD, ABD, the angle CGH will be equal to the inclination of these planes (6. def. 11.); that is to the spherical angle ABC. The sine therefore of the hypothenuse CB is to the radius, as the sine of the side AC is to the sine of the opposite angle ABC. Q. E. D.

COR. Of these three, viz. the hypothenuse, a side, and the angle opposite to that side, any two being given, the third is also given by prop. 2.

PROP. XIX FIG. 14.

IN right angled spherical triangles, the co-sine of the hypothenuse is to the radius, as the co-tangent of either of the angles is to the tangent of the remaining angle.

Let ABC be a spherical triangle, having a right angle at A; the co-sine of the hypothenuse BC will be to the radius, as the co-tangent of the angle ABC to the tangent of the angle ACB.

Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through the pole B of the circle DF, DF will also pass through the pole BD (13. 18. 1. Theod. sph.). And since AC is perpendicular to BD, AC will also pass through the pole of BD; wherefore the pole of the circle BD will be found in the point where the circles AC, DE meet, that is, in the point F: the arches FA, EF are therefore quadrants, and likewise the arches BD, BE: in the triangle CEF, right angled at the point E, CE is the complement of the hypothenuse BC of the triangle ABC, BF is the complement of the arch ED, which is the measure of the angle ABC, and FC the hypothenuse of the triangle CEF, is the complement of AC; and the arch AD, which is the measure of the angle CFE, is the complement of AB.

But (17. of this) in the triangle CEF, the sine of the side CE is to the radius, as the tangent of the other side is to the tangent of the angle ECF opposite to it, that is, in the triangle ABC, the co-sine of the hypothenuse BC is to the radius, as the co-tangent of the angle ABC is to the tangent of the angle ACB. Q. E. D.

COR. 1. Of these three, viz. the hypothenuse and the two angles, any two being given, the third will also be given.

COR. 2. And since by this proposition the co-sine of the hypothenuse BC is to the radius as the co-tangent of the angle ABC to the tangent of the angle ACB; but as the radius is to the co-tangent of the angle ACB, so is the tangent of the same to the radius (Cor. 2. def. Pl. Tr.); and ex æquo, the co-sine of the hypothenuse BC is to the co-tangent of the angle ACB, as the co-tangent of the angle ABC to the radius.

PROP. XX. FIG. 14.

IN right angled spherical triangles, the co-sine of an angle is to the radius, as the tangent of the side adjacent to that angle is to the tangent of the hypothenuse.

The same construction remaining; in the triangle CEF (17. of this), the sine of the side EF is to the radius, as the tangent of the other side CE is to the tangent of the angle CFE opposite to it; that is, in the triangle ABC, the co-sine of the angle ABC is to the radius, as (the co-tangent of the hypothenuse BC to the co-tangent of the side AB, adjacent to ABC, or as) the tangent of the side AB to the tangent of the hypothenuse, since the tangents of two arches are reciprocally proportional to their co-tangents. (Cor. 1 def. Pl. Tr.)

COR. And since by this proposition the co-sine of the angle ABC is to the radius, as the tangent of the side AB is to the tangent of the hypothenuse BC; and as the radius is to the co-tangent of BC, so is the tangent of BC to the radius; by equality, the co-sine of the angle ABC will be to the co-tangent of the hypothenuse BC, as the tangent of the side AB, adjacent to the angle ABC, to the radius.

PROP. XXI. FIG. 14.

IN right angled spherical triangles, the co-sine of either of the sides is to the radius, as the co-sine of the hypothenuse is to the co-sine of the other side.

The same construction remaining; in the triangle CEF, the sine of the hypothenuse CF is to the radius, as the sine of the side CE to the sine of the opposite angle CFE (18. of this); that is, in the triangle ABC, the co-sine of the side CA is to the radius, as the co-sine of the hypothenuse BC to the co-sine of the other side BA. Q. E. D.

PROP. XXII. FIG. 14.

IN right angled spherical triangles, the co-sine of either of the sides is to the radius, as the co-sine of the angle opposite to that. side is to the sine of the other angle.

The same construction remaining; in the triangle CEF, the sine of the hypothenuse CF is to the radius, as the sine of the side EF is to the sine of the angle ECF opposite to it; that is, in the triangle ABC, the co-sine of the side CA is to the radius, as the co-sine of the angle ABC opposite to it, is to the sine of the other angle. Q. E. D.

OF THE CIRCULAR PARTS.

Fig. 15.-IN any right angled spherical triangle ABC, the complement of the hypothenuse, the complements of the angles, and the two sides, are called The circular parts of the triangle, as if it were following each other in a circular order, from whatever part we begin thus, if we begin at the complement of the hypothenuse, and proceed towards the side BA, the parts following in order will be the complement of the hypothenuse, the complement of the angle B, the side BA, the side AC, (for the right angle at A is not reckoned among the parts,) and, lastly, the complement of the angle C. And thus at whatever part we begin, if any three of these five be taken, they either will be all contiguous or adjacent, or one of them will not be contiguous to either of the other two: in the first case, the part which is between the other two is called the Middle part, and the other two are called Adjacent extremes. In the second case, the part which is not contiguous to either of the other two is called the Middle part, and the other two Opposite extremes. For example, if the three parts be the complement of the hypothenuse BC, the complement of the angle B, and the side BA; since these three are contiguous to each other, the complement of the angle B will be the middle part, and the complement of the hypothenuse BC and the side BA will be adjacent extremes: but if the complement of the hypothenuse BC, and the sides BA, AC be taken; since the complement of the hypothenuse is not adjacent to either of the sides, viz. on account of the complements of the two angles B and C intervening between it and the sides, the complement of the hypothenuse BC will be the middle part, and the sides BA, AC opposite extremes. The most acute and ingenious Baron Napier, the inventor of Logarithms, contrived the two following rules concerning these parts, by means of which all the cases of right angled spherical triangles are resolved with the greatest ease.

RULE I.

The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the tangents of the adjacent parts.

RULE II.

The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the co-sines of the opposite parts.

These rules are demonstrated in the following manner:

Fig. 15. First. Let either of the sides, as BA, be the middle part, and therefore the complement of the angle B, and the side AC will be adjacent extremes. And by Cor. 2. prop. 17, of this, S, BA is to the Co-T, B, as T, AC is to the radius, and therefore B x S, BA Co.T, B x T, AC.

The same side BA, being the middle part, the complement of the hypothenuse, and the complement of the angle C, are opposite