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sphere, a solid angle may be conceived as formed at O, by the plane angles AOB, BOC, COD, &c.; and the sum of these angles must be less than four right angles; which is exactly the proposition we have been engaged with. The demonstration here given is different from that of Prop. XXII B. I. El. S. Geom.; both, however, suppose that the polygon ABCDE is convex, or that no side produced will cut the figure.

PROPOSITION VI. THEOREM.

The poles of a great circle of the sphere, are the extremities of that diameter of the sphere which is perpendicular to this circle; and these extremities are also the poles of all small circles parallel to it.

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wise are all the arcs EA, EM, EB, &c.; hence the points D and E are each equally distant from all the points of the circumference AMB; hence (Def. 5.) they are the poles of that circumference.

Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence (Prop. I. Cor. 4.) it passes through O the centre of the circle FNG; hence, if the oblique lines DF, DN, DG be drawn, these oblique lines will diverge equally from the perpendicular DO, and will themselves be equal. But, the chords being equal, the arcs are equal; hence the point D is the pole of the small circle FNG; and, for like reasons, the point É is the other pole.

Cor. 1. Every arc DM, drawn from a point in the arc of a great circle AMB to its pole, is a quarter of the circumfer.

ence, which, for the sake of brevity, is usually named a quadrans, or quadrant: and this quadrant at the same time makes a right angle with the arc AM. For (Prop. XVII. B. I. El. S. Geom.) the line DC being perpendicular to the plane AMC, every plane DMC passing through the line DC, is perpendicular to the plane AMC; hence the angle of these planes, or (Def. 6.) the angle AMD, is a right angle.

Cor. 2. To find the pole of a given arc AM, draw the indefinite arc MD perpendicular to AM; take MD equal to a quadrant; the point D will be one of the poles of the arc AM: or thus, at the two points A and M, draw the arcs AD and MD perpendicular to AM; their point of intersection, D, will be the pole required.

Cor. 3. Conversely, if the distance of the point D from each of the points A and M, is equal to a quadrant, the point D will be the pole of the arc AM, and also the angles DAM, AMD, will be right.

For, let C be the centre of the sphere, and draw the radii CA, CD, CM. Since the angles ACD, MCD are right, the line CD is perpendicular to the two straight lines CĂ, CM; hence it is perpendicular to their plane (Prop. V. B. I. El. S. Geom.); hence the point D is the pole of the arc AM; and consequently the angles DAM, AMD are right.

Scholium. The properties of these poles enable us to describe arcs of a circle on the surface of a sphere, with the same facility as on a plane surface. It is evident, for instance, that by turning the arc DF, or any other line extending to the same distance, round the point D, the extremity F will describe the small circle FNG; and, by turning the quadrant DFA round the point D, its extremity A will describe the are of the great circle AM.

If the arc AM were required to be produced, and nothing were given but the points A and M through which it was to pass, we should first have to determine the pole D, by the intersection of two arcs described from the points A and M as centres, with a distance equal to a quadrant. The pole D being found, we might describe the arc AM and its prolongation, from D as a centre, and with the same distance as before.

In fine, if it is required from a given point P to let fall a perpendicular on the given arc AM, produce this arc to S, till the distance PS be equal to a quadrant; then, from the pole S, and with the same distance, describe the arc PM, which will be the perpendicular required.

PROPOSITION VII. THEOREM.

Every plane perpendicular to a radius at its extremity is tangent to the sphere.

Let FAG (see the next diagram) be a plane perpendicular to the radius OÀ, at its extremity A. Any point M in this plane being assumed, and OM, AM being joined, the angle ŎAM will be right, and hence the distance OM will be greater than OA. Hence the point M lies without the sphere; and as the same can be shown for every other point of the plane FAG, this plane can have no point but A common to it and the surface of the sphere; hence (Def. 4.) it is tangent.

Scholium. In the same way, it may be shown that two spheres have but one point in common, and therefore touch each other, when the distance between their centres is equal to the sum, or the difference of their radii; in which case, the centres and the point of contact lie in the same straight line.

PROPOSITION VIII. THEOREM.

The angle formed by two arcs of great circles, is equal to the angle formed by the tangents of these two arcs at their points of intersection, and is measured by the arc described from this point of intersection, as a pole, and limited by the sides, produced if necessary.

Let the angle BAC be formed by the two arcs AB, AC; then it will be equal to the angle FAG formed by the tangents AF, AG, and be measured by the arc DE, described about A as a pole.

For the tangent AF, drawn in the plane of the arc AB, is perpendicular to the radius AO; and the tangent AG, drawn in the plane of the arc AC, is perpendicular to the same radius AO. Hence, (Prop. XX. B. I. El. S. Geom.) the angle FAG is equal to the angle contained by the planes OAB, OAC; which is that of the arcs AB, AC, and is named BAC.

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In like manner, if the arcs AD and AE are both quadrants, the line OD, OE will be perpendicular to OA, and the angle DOE will still be equal to the angle of the planes AOD, AOE: hence the arc DE is the measure of the angle contained by these planes, or of the angle CAB.

Cor. The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles and included between their sides: hence it is easy to make an angle of this kind equal to a given angle.

Scholium. Vertical angles, such as ACO and BCN (see diagram to Prop. XXI.) are equal; for either of them is still the angle formed by the two planes ACB, OCN.

It is further evident, that, in the intersection of two arcs ACB, OCN, the two adjacent angles ACO, OCB taken together, are equal to two right angles.

PROPOSITION IX. THEOREM.

If from the vertices of the three angles of a spherical triangle, as poles, three arcs be described forming a second triangle, the vertices of the angles of this second triangle will be respectively poles of the sides of the first.

From the vertices A, B, C, as poles, let the arcs EF, FD, ED be described, forming on the surface of the sphere, the triangle DFE; then will the points D, E, and F be respectively poles of the sides BC, AC, AB.

For, the point A being the pole of the arc EF, the distance AE is a quadrant; the

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point C being the pole of the arc DE, the distance CE is likewise a quadrant: hence the point E is removed the length of a quadrant from each of the points A and C ; hence (Prop VI. Cor. 3.) is the pole of the arc AC. It might be shown, by the same method, that D is the pole of the arc BC, and F that of the arc AB.

Cor. Hence the triangle ABC may be described by means of DEF, as DEF is described by means of ABC.

PROPOSITION X. THEOREM.

The same supposition continuing as in the last proposition, each angle in the one of the triangles, will be measured by the semicircumference minus the side lying opposite to it in the other triangle.

Produce the sides (see the preceding diagram) AB, AC, if

necessary, till they meet EF, in G and H. The point A being the pole of the arc GH, the angle A will be measured by that arc. But the arc EH is a quadrant, and likewise GF, E being the pole of AH, and F of AG; hence EH+GF is equal to the semicircumference. Now, EH+GF is the same as EF+GH; hence the arc GH, which measures the angle A, is equal to a semicircumference minus the side EF. In like manner, the angle B will be measured by circ.-DF: the angle C, by circ.-DE.

And this property must be reciprocal in the two triangles, since each of them is described in a similar manner by means of the other. Thus we shall find the angles D, E, F, of the triangle DEF to be measured respectively by circ.-BC, circ.-AC. circ.-AB. Thus the angle D, for example, is measured by the arc MI; but MI+BC=MC+BI=1 circ. ; hence the arc MI, the measure of D, is equal to circ.—BC; and so of all the rest.

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Scholium. It must further be observed, that besides the triangle DEF, three others might be formed by the intersection of the three arcs DE, EF, DF. But the proposition immediately before us is applicable only to the central triangle, which is distinguished from the other three by the circumstance (see diagram to Prop. IX.) that the two angles A and D lie on the same side of BC, the two B and E on the same side of AC, and the two C and F on the same side of AB.

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Various names have been given to the triangles ABC, DEF; we shall call them polar triangles.

PROPOSITION XI. THEOREM.

If around the vertices of the two angles of a given spherical triangle, as poles, the circumference of two circles be described which shall pass through the third angle of the triangle; if then, through the other point in which those circumferences intersect, and the two first angles of the triangle, the arcs of great circles be drawn, the triangle thus formed will have all its parts equal to those of the first triangle.

Let ABC be the given triangle, CED, DFC the arc described about A and B as poles; then will the triangle ADB have all its parts equal to those of ABC.

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