Page images

Scholium. This proposition is fundamentally the same as (Book VI. Prop. XX.); for, O being the centre of the sphere, a solid angle may be conceived as formed at Oby 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 here proved. The

[blocks in formation]

demonstration here given is different from that of Book VI. Prop. XX.; both, however, suppose that the polygon ABCDE is convex, or that no side produced will cut the figure.


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

Let ED be perpendicular to the great circle AMB; then will E and D be its poles; as also the poles of the parallel small circles HPİ, FNG.

For, DC being perpendicular to the plane A AMB, is perpendicular to all the straight lines CA, CM, CB, &c. drawn through its foot in this plane; hence all the arcs DA, DM, DB, &c. are quarters of the circumference. So likewise are

[merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

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, they are the poles of that circumference (Def. 7.). Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence, it passes through O the centre of the circle FNG (Book VIII. Prop. VII. Cor. 4.); 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 E 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 circumference, which for the sake of brevity, is usually named a quadrant: and this quadrant at the same time makes a right angle with the arc AM. For, the line DC being perpendicular to the plane AMC, every plane DME, passing through the line DC is perpendicular to

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

the plane AMC (Book VI. Prop. XVI.); hence, the angle of these planes, or 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 angles.

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

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 arc of the great circle AMB.

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 be required from a given point P, to let fall a perpendicular on the given arc AM; find a point on the arc AM at a quadrant's distance from the point P, which is done by describing an arc with the point P as a pole, intersecting AM in S: S will be the point required, and is the pole with which a perpendicular to AM may be described passing through the point P.


The angle formed by two arcs of great circles, is equal to the ungle formed by the tangents of these arcs at their point 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 A arcs AB, AC; then will it 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 the angle FAG is equal to the angle contained by the planes ABO, OAC (Book VI. Def. 4.); which is that of H the arcs AB, AC, and is called the angle BAC.

[blocks in formation]

In like manner, if the arcs AD and AE are both quadrants, the lines OD, OE, will be perpendicular to OA, and the angle DOE will still be equal to the angle of the planes AOD, AOË : 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 are equal; for either of them is still the angle formed by the two planes ACB, OCN.

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





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

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

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, it is the pole of the arc AC (Prop. V. Cor. 3.). 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. Triangles so described are called polar triangles, or supplemental tri angles


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








For, produce the sides 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 (Prop. VI.). 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 a 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.


[ocr errors]


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={ circ.; hence the arc MI, the measure of D, is equal to circ.-BC; and so of all the rest.

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 the last figure) 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.


« PreviousContinue »