H its altitude: the solidity of the truncated cone will be π.H. (A2+ B2+AB).

Let R be the radius of a sphere; its solidity will be R3. Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base: the solidity of the sector will be RH.

Let P and Q be the two bases of a spherical segment, H its P+Q.H+.H3.

altitude: the solidity of the segment will be 2

If the spherical segment has but one base, the other being nothing, its solidity will be PH+дπН3.

Definitions.

1. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.

These arcs are named the sides of the triangle, and are always supposed to be each less than a semi-circumference. The angles, which their planes form with each other, are the angles of the triangle.

2. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 3. A spherical polygon is a portion of the surface of a sphere terminated by several arcs of great circles.

4. A lune is that portion of the surface of a sphere, which is included between two great semi-circles meeting in a common diameter.

5. A spherical wedge or ungula is that portion of the solid sphere, which is included between the same great semi-circles, and has the lune for its base.

6. A spherical pyramid is a portion of the solid sphere, included between the planes of a solid angle whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes.

7. The pole of a circle of a sphere is a point in the surface equally distant from all the points in the circumference of this circle. It will be shown (Prop. V.) that every circle, great or small, has always two poles.

PROPOSITION I. THEOREM.

In every spherical triangle, any side is less than the sum of the

other two.

Let O be the centre of the sphere, and ACB the triangle; draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be drawn; these planes will form a solid angle at the centre O; and the angles AOB, AOC, COB, will be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid angle is less than the sum of the other two (Book VI. Prop. XIX.); hence any side of the triangle ABC is less than the sum of the other two.

A

Let ANB be the arc of a great circle which joins the points A and B; then will it be the shortest path between them.

1st. If two points N and B, be taken on the arc of a great circle, at unequal distances from the point A, the shortest distance from B to A will be greater than the shortest distance from N to A.

PROPOSITION II. THEOREM.

The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points.

M

B

NP D+

B

For, about A as a pole describe a circumference CNP. Now, the line of shortest distance from B to A must cross this circumference at some point as P. But the shortest distance from P to A whether it be the arc of a great circle or any other line, is equal to the shortest distance from N to A; for, by passing the arc of a great circle through P and A, and revolving it about the diameter passing through A, the point P may be made to coincide with N, when the shortest distance from P to A will coincide with the shortest distance from N to A: hence, the shortest distance from B to A, will be greater than the shortest distance from N to A, by the shortest distance from B to P.

If the point B be taken without the arc AN, still making AB greater than AN, it may be proved in a manner entirely similar to the above, that the shortest distance from B to A will be greater than the shortest distance from N to A.

If now, there be a shorter path between the points B and A, than the arc BDA of a great circle, let M be a point of the short

est distance possible; then through M draw MA, MB, arcs of great circles, and take BD equal to BM. By the last theorem, BDA<BM+MA; take BD=BM from each, and there will remain AD<AM. Now, since BM=BD, the shortest path from B to M is equal to the shortest path from B to D: hence if we suppose two paths from B to A, one passing through M and the other through D, they will have an equal part in each; viz. the part from B to M equal to the part from B to D.

But by hypothesis, the path through M is the shortest path from B to A: hence the shortest path from M to A must be less than the shortest path from D to A, whereas it is greater since the arc MA is greater than DA; hence, no point of the shortest distance between B and A can lie out of the arc of the great circle BDA.

PROPOSITION III. THEOREM.

The sum of the three sides of a spherical triangle is less than the circumference of a great circle,

Let ABC be any spherical triangle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be semicircumferences, since two great circles always bisect each other (Book VIII, Prop. VII. A Cor. 2.). But in the triangle BCD, we have the side BC<BD+CD (Prop I.); add AB+ AC to both; we shall have AB+AC+BC<ABD+ACD, thatisto say,less than a circumference.

B

D

E

PROPOSITION IV. THEOREM

The sum of all the sides of any spherical polygon is less than the circumference of a great circle,

E

C

Take the pentagon ABCDE, for example. Produce the sides AB, DC, till they meet in F; then since BC is less than BF+CF, the perimeter of the pentagon ABCDE will be less than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till they meet in G; we shall have ED<EG+DG; hence the pe rimeter of the quadrilateral AEDF is less than that of the triangle AFG; which last is itself less than the circumference of a great circle; hence, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference.

A

B

E

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 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 here proved. The 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.

A

E

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 the plane AMC (Book VI. Prop. XVI.); hence, the angle of these planes, or the angle AMD, is a right angle.

E

H

M M

Q

T

B

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