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1. Let CD fall within the triangle ; then, since ADC, BDC are right angled spherical triangles, the angles A, B must each be of the same affection with CD, (14).

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D A 2. Let CD fall without the triangle ; then (14.) the angle B is of the same affection with CD ; and the angle CAD is of the same affection with CD ; therefore the angle CAD and B are of the same affection, and the angle CAB and B are therefore of different affections. Q. E. D.

Cor. Hence, if the angles A and B be of the same affection, the perpendicular will fall within the base ; for if it did not, A and B would be of different affection. And if the angles A and B be of different affection, the perpendicular will fall without the triangle ; for, if it did not, the angles A and B would be of the same affection, contrary to the supposition.

PROP. XVII. If to the base of a spherical triangle a perpendicular be drawn from the

opposite angle, which either falls within the triangle, or is the nearest of the two that fall zithout; the least of the segments of the base is adjacent to the least of the sides of the triangle, or to the greatest, according as the sum of the sides is less or greater than a semicircle.

Let ABEF be a great circle of a sphere, H its pole, and GHD any circle passing through H, which therefore is perpendicular to the circle ABEF. Let A and B be two points in the circle ABEF, on opposite sides of the point D, and let D be nearer to A than to B, and let C be any point in the circle GHD

B В between H and D. Through the points A and C, B and C, let the arches AC and BC be drawn, and let them be produced till they meet the circle ABEF in the points E and F, then the arches ACE, BCF are semicircles. Also ACB, ACF, CFE, ECB are four spherical triangles contained by arches of the same

F circles, and having the same perpendiculars CD and CG.

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1. Now because CE is nearer to the arch CHG than CB is, CE is greater than CB, and therefore CE and CA are greater than CB and CA, wherefore CB and CA are less than a semicircle ; but because AD is by supposition less than DB, AC is also less than CB, (13.), and therefore in this case, viz. when the perpendicular falls within the triangle, and when the sum of the sides is less than a semicircle, the least segment is adjacent to the least side.

2. Again, in the triangle FCA the two sides FC and CA are less than a semicircle ; for, since AC is less than CB, AC and CF are less than BC and CF. Also, AC is less than CF, because it is more remote from CHG than CF is ; therefore in this case also, viz. when the perpendicular falls without the triangle, and when the sum of the sides is less than a semicircle, the least segment of the base AD is adjacent to the least side.

3. But in the triangle FCE the two sides FC and CE are greater than a semicircle ; for, since FC is greater than CA, FC and CE are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greater side.

4. In the triangle ECB the two sides EC, CB are greater than a semicircle ; for, since by supposition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB, wherefore in this case, also, the least segment of the base EG is adjacent to the greatest side of the triangle. Therefore, when the sum of the sides is greater than a semicircle, the least segment of the base is adjacent to the greatest side, whether the perpendicular fall within or without the triangle : and it has been shewn, that when the sum of the sides is less than a semicircle, the least segment of the base is adjacent to the least of the sides, whether the perpendicular fall within or without the triangle. Wherefore, &c. Q. E. D.

PROP. XVIII.

In right angled sphérical 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 AF be drawn perpendicular to BD, which therefore will be the size of the arch AB, and from the point F; let there be drawn in the pkane BDC the straight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the straight

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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.); where. fore 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,

B (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.

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 ra

dius 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

C 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 com

D. mon section of two planes, DCF,

A ECF, each perpendicular to the plane ADB. Wherefore CFD, CFE are right angles, and CF is the sine

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

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, liaving 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

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A the połe 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 CE : R : : 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 ABC : R:: tan AB : tan BC. Therefore, &c. Q. E. D.

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А. 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 CA : R :: cos BC : cos AB. Q. £. D.

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