line DF is at right angles to both FA and FE, it will also be at right angles to the plane AEF (4. 2. Sup.); wherefore the plane ABD, which passes through DF is perpendicular to the plane AEF (17. 2. Sup.), and the plane AEF perpendicular to ABD: But the plane ACD or AED, is also perpendicular to the same ABD, because the spherical angle BAC is a right angle: Therefore AE, the common section of the planes AED, AEF, is at right angles to the plane ABD, (18. 2. Sup.), and EAF, EAD are right angles. Therefore AE is the tangent of the arch AC; and in the rectilineal triangle AEF, having a right angle at A, AF is to the radius as AE to the tangent of the angle AFE, (1. Pl. Tr.); but AF is the sine of the arch AB, and AE the tangent of the arch AC; and the angle AFE is the inclination of the planes CBD, ABD, (4. def. 2. Sup.), or is equal to 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. Q. E. D.

F

B

COR. Since by this proposition, sin AB : R:: tan AC : tan ABC ; and because R cot ABC tan ABC R (1. Cor. def. 9, Pl. Tr.) by equality, sin AB cot ABC; tan AC : R.

PROP. XIX.

In right angled spherical triangles the sine of the hypotenuse is to the radius us 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 sign of the hypotenuse 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 CE be drawn perpendicular to DB, which will therefore be the sine of the hypotenuse BC; and from the point E let there be drawn in the plane ABD the straight line EF perpendicular to DB, and let CF be joined then CF will be at right angles to the plane ABD, because as was shown of EA in the preceding proposition, it is the common section of two planes, DCF, ECF, each perpendicular to the plane ADB. Wherefore CFD, CFE are right angles, and CF is the sine

D

E

of the arch AC; and in the triangle CFE having the right angle CFE;

CE is to the radius, as CF to the sine of the angle CEF (1. Pl. Tr). But, since CE, FE are at right angles to DEB, which is the common section of the planes CBD, ABD, the angle CEF is equal to the inclination of these planes, (4. def. 2. Sup.), that is to the spherical angle ABC. Therefore the sine of the hypotenuse CB, is to the radius, as the sine of the side AC to the sine of the opposite angle ABC. Q. E. D.

PROP. XX.

In right angled spherical triangles, the cosine of the hypotenuse is to the radius as the cotangent 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 cosine of the hypotenuse BC is to the radius as the cotangent 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 must pass through the pole of BD, (4.) And since AC is perpendicular to BD, the plane of the circle AC is perpendicular to the plane of the circle BAD, and therefore AC must also (4.) pass through the pole of BAD; wherefore, the pole of the circle BAD is in the point F, where the circles AC, DE, intersect. The arches FA, FD are therefore quadrants, and likewise the arches BD, BE. Therefore, in the triangle CEF, right angled at the point E, CE is the complement of BC, the hypotenuse of the triangle ABC; EF is the complement of the arch ED, the measure of the angle ABC, and FC, the hypotenuse 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 (18.) in the triangle CEF, sin CER: tan EF: tan ECF, that is, in the triangle ACB, cos BC: R:: cot ABC ; tan ACB. Q. E. D.

COR. Because cos BC: R:: cot ABC : tan ACB, and (Cor. 1, def. 9, Pl. Tr.) cot ACB: R :: R: tan ACB, ex æquo, cot ACB : cos BC : : R: cot ABC.

PROP. XXI.

In right angled spherical triangles, the cosine of an angle is to the radius as the tangent of the side adjacent to that angle is to the tangent of the hypotenuse.

=

The same construction remaining: In the triangle CEF, sin FE : R: tan CE tan CFE (18.); but sin EF cos ABC; tan CE= cot BC, and tan CFE cot AB, therefore cos ABC: R cot BC: cot AB. Now, because (Cor. 1. def. 9. Pl. Tr.) cot BC: R:: R : tan BC, and cot AB R:: R: tan AB, by equality inversely, cot BC : cot AB tan AB tan BC; therefore (11. 5.) cos ABČ: R:: tan AB: tan BC. Therefore, &c. Q. E. D.

COR. 1. From the demonstration it is manifest, that the tangents of any two arches AB, BC are reciprocally proportional to their cotangents.

Cor. 2. Because cos ABC: R:: tan AB : tan BC, and R : cot BC :: tan BC: R, by equality, cos ABC: cot BC :: tan AB: R. That is, the cosine of any of the oblique angles is to the cotangent of the hypotenuse, as the tangent of the side adjacent to the angle is to the radius.

PROP. XXII.

In right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the hypotenuse is to the cosine of the other side.

The same construction remaining: In the triangle CEF, sin CF : R:: sin CE sin CFE, (19.); but sin CF cos CA, sin CE = cos BC, and sin CFE cos AB; therefore, cos CAR:: cos BC: cos AB. Q. E. D.

PROP. XXIII.

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

The same construction remaining: In the triangle CEF, sin CF : R sin EF sin ECF, (19.); but sin CF = cos CA, sin EF = cos ABC, and sin ECF-sin BCA; therefore, cos CA: R cos ABC: sin BCA. Q. E. D.

PROP. XXIV.

In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them.

First, Let ABC be a right angled triangle, having a right angle at A; therefore, (19.) the sine of the hypotenuse BC is to the radius, (or the sine of the right angle at A), as the sine of the side AC to the sine of the angle B. And, in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C.

A

Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the sine of the angle A opposite to BC, is to the sine of the angle B opposite to AC. Through the point C, let there be drawn an arch of a great circle CD perpendicular to AB; and in the right angled trian

gle BCD, sin BC: R :: sin CD: sin B, (19.); and in the triangle ADC, sin AC: R:: sin CD : sin A; wherefore, by equality inversely, sin BC sin AC :: sin A: sin B. In the same manner, it may be proved that sin BC: sin AB sin A: sin C, &c. Therefore, &c. Q. E. D.

PROP. XXV.

In oblique angled spherical triangles, a perpendicular arch being drawn from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle.

Let ABC be a triangle, and the arch CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD.

For having drawn CD perpendicular to AB, in the right angled triangle BCD, (23.) cos CD: R:: cos B: sin DCB; and in the right angled triangle ACD, cos CD: R:: cos A: sin ACD; therefore (11.5.) cos B: sin DCB :: cos A: sin ACD, and alternately, cos B : cos A sin BCD : sin ACD. Q. E. D.

PROP. XXVI.

The same things remaining, the cosines of the sides BC, CA, are propor tional to the cosines of BD, DA, the segments of the base.

For in the triangle BCD, (22.), cos BC: cos BD :: cos DC: R, and in the triangle ACD, cos AC: cos AD: : cos DC: R; therefore (11. 5.) cos BC: cos BD :: cos AC : cos AD, and alternately, cos BC cos AC :: cos BD : cos AD. Q. E. D.

PROP. XXVII.

The same construction remaining, the sines of BD, DA the segments of the base are reciprocally proportional to the tangents of B and A, the angles at the base.

In the triangle BCD, (18.), sin BD: R; tan DC : tan B ; and in the triangle ACD, sin AD: R :: tan DC: tan A; therefore, by equality inversely, sin BD: sin AD :: tan A: tan B. QE. D.

AD

B D

B