53. IMAGINE that the semi-circle PAp revolves P round its diameter Pp; and by its revolution de- d. scribes the sphere APAp; it is evident that the A radius AC will describe a great circle, while the lines bd, ef, will describe smaller circles, di

r minishing as they recede from the radius AC.

54. Any other great semi-circle equal to the first, the semicircle APA, for example, would have generated the same sphere by its revolution round the diameter AA. Hence, since there are an infinite number of equal diameters about which this revolution may take place, there are evidently an infinite, number of great circles in any spheres, all of which are equal.

There are alse an infinite number of smaller circles; but because of their inequality they are not used in Spherical Trigono metry.

55. The name of SPHERICAL TRIGONOMETRY is given to that science which instructs us to resolve triangles formed upon the surface of a sphere by three of its great circles.

Take any spherical body, a billiard ball for example, trace upon it three circular arcs whose planes pass through its center; these arcs, by their intersection, will form a spherical triangle, of which they will be the sides.

Each of these arcs belong to a circle whose center is also the center of the ball, and through this center we may suppose that a diameter of the ball passes perpendicularly to the plane of the circle.

56. A diameter thus perpendicular to the plane of a great circle, is called the axis of this circle. The two extremities of the axis are called the poles of the circle. Hence the points P, p are the poles of the circle ACA.

Each great circle has therefore its own two poles, since two great cireles cannot have the same axis.

57. And since the axis of a great circle must always be perpendicular to its plane, and pass through its centre, it evidently follows

1st. That it forms as many right angles as there are radii in the plane of the circle.

2nd. That the arcs which measure these angles are all 90°, and therefore equal, since they belong to equal circles. Hence their chords are equal.

These chords measure the distance from the pole of the circle to

the different points of its circumference. Therefore the pole of any circle is equally distant from every point of its circumference.

58. Thus we may say that the pole of a circle is a point in the surface of the sphere 90o distant from every point of the circumference of this circle; or which amounts to the same thing, that the arc contained between the pole of a oircle and any point of its circumference, is always 90°.

Hence it is very easy to trace on the surface of a sphere, a circle whose poles are known; and reciprocally to find the poles of a given circle.

The spherical compasses solve this double problem with the greatest facility.

59. Since all great circles of the sphere have a common center, the line of their intersection is necessarily a diameter of the sphere, which is at the same time the diameter of all these circles. But every diameter bisects its circle, therefore two or more great circles divide one another into equal parts.

And consequently,

1st. If two great circles have already intersected each other, they will again intersect each other at 180° from the first point of inter section, and not sooner.

2nd. Hence it is not possible to include between only two arcs, even the smallest portion of a spherical surface, unless they are each 180.

Whatever be the number of degrees which the two arcs contain, they form on the surface of the sphere an angle, of which it is important to ascertain the measure ; and this we shall proceed to in. vestigate.

60. Let there be two arcs AB and AD, each of which I suppose to contain 90°, and which by their concurrence at A, form a spherical angle BAD. It is clear

Ist. That the radii BC, DC, are both perpendicular to AC.

2nd. That these two radii have the same inclination towards each other, as the circular planes to which they belong.

Srd. That the measure of this inclination is the arc BD, since the angle BCD has its summit at the center.

61. Hence the measure of any spherical angle is the arc of a great circle comprised between its sides at goo distance from the summit of that angle.

In general, if from the summit of a spherical angle taken as a pole, we describe an arc of a great circle, the portion of this arc comprised between the sides of the angle, will always be its measure.

Not but that it would be possible to measure the angle by the arcs of small circles ; for it is easily perceived that the angle bod, for example, formed by the sines of the arcs Ab, Ad, is equal to the angle BCD; and consequently that the arc bd, which mea


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sures it, is of the same number of degrees as the arc BD. But for the sake of uniformity, it is usual to measure the value of spherical angles upon great circles described from their summits as poles.

62. Hence spherical angles are always equal to those which at 90° distance from their summits are formed by the sines of the arcs composing their sides. But any angle formed by tuo sines must be less than 180°; therefore every spherical angle is less than 180°.

63. Since spherical angles are equal to the angles formed by the sines of their sides, it evidently follows

1st. That when one arc of a great circle falls upon another, the angles which result are equal to two right angles.

2nd. That if we prolong these two arcs beyond their point of intersection, the vertical angles are equal, as well as those formed by their sines, if prolonged in a similar manner.

3rd. That the sum of the angles formed about the point of intersection, is 360°.

64. Suppose now that the three circles, of which the sides of a spherical triangle are a part, be entirely described; it is clear that they will form on the other hemisphere a triangle perfectly equal to the first. It is also evident that, drawing the respective chords of all the arcs which form these two spherical triangles, there will result two rectilinear triangles perfectly equal, the one above, and the other below the center of the sphere.

But we know that in any right-lined triangle the sum of any two sides is greater than the third side. Therefore in any spherical triangle A ABC, the sum of any two arcs is also “greater than the third arc or side.

Again since a right line is the shortest line that can be drawn from one point to another, it follows that the arc of a great circle passing through two points of the surface of a sphere, is the shortest line that can be drawn between those points. Hence it measures the distance between them.

65. We have already said (59) that it was not possible to enclose on all sides between only two arcs, any portion of a spherical surface, unless each arc was 180°.

Therefore every spherical triangle CAB, results from the intersection of two arcs, cut by a third, before these arcs re-unite. But they meet again only at 180° distance from the point where they first intersected one another; therefore

А side of any

A a spherical triangle is less than 180°.

66. If we prolong the two arcs CA and CB of the triangle CAB, till they again meet in D, it is evident that AB will be less than the sum

of the prolongations AD, DB. But this sum added to the two prolonged arcs, is only 360o.


Therefore the sum of the sides of any spherical triangle is alway. less than 360o.

67. As to the sum of the angles, it can never equal six right angles, since each angle must be less than 180°. This sum has therefore a limit which it cannot exceed, which limit is 540°. Has it not also a limit which it cannot fall short of? This we shall

proceed to determine.

68. From the three angles ABC, taken successively for poles, describe the arcs EF, DF, and DE, which by their concourse will form the exterior triangle FDE.

This construction shews us,

ist. That the point A is 90' distant from every point of the arc EF;

2nd. That the point B is similarly 90° from every point of the arc DF; therefore F is the pole of the arc AB.

In the same manner we prove that the points E and D are respectively the poles D of the arcs CA, CB.


E This premised, prolong the two arcs CA, CB, till they meet the arc DE: we shall have EG-90°, as well as DH; therefore EG +DH, or which is the same, EG+ DG+GH, or ED+GH=180° The arc ED is therefore the supplement of the arc GH, and consequently of the angle C, of which GH is the measure (61).

In the same manner we prove that che arcs EF, DF, are the respective supplements of the angles A and B.

Continue now the arc GC till it meets the arc EF; we shall find that GI will be the measure of the angle E, and that the part AC will be its supplement, since GC+Al, or GI + AC=180°.

In like manner the arc AB will be the supplement of the angle F;

and the arc BC, that of the angle D.

Whence in general we may conclude that, if from the three angles of any spherical triangle, taken as poles, we deseribe titree arcs. whose meeting forms a neto spherical triangle, the angles and sides of this second triangle will be reciprocally the sæpplements of the sides and angles opposite to them in the first triangle.

Thus the angle E of the exterior triangle + the arc AC, opposite to it in the interior triangle, measure 180°; and recipro cally the arc DE of the exterior triangle + the angle C, which is opposite to it in the interior triangle, also measure 180o.

69. This exterior triangle is called the supplemental triangle. It is of great use in Spherical Trigonometry. We shall at present employ it to discover the least limit of the value of the three angles ABC.

These three angles have for their respective supplements the three sides of the supplemental triangle. With them they form the sum of six right angles. But the sum of those three sides is always less than four right angles (66). Therefore the sum of the three angles ABC is necessarily greater than 180°.


70. From what has been stated, it follows :

ist. That the sum of the angles of a spherical triangle may vary from 180° to 540° exclusively. And consequently we cannot infer the value of the third angle from the other two, as in plane triangles.

2ly. That the three angles of a spherical triangle may be right, or even obtuse, as well as acute, provided that in this last case their sum exceeds 1800.

71. If we desire to judge of the relation between two spherical triangles, whose angles are respectively equal, we must have recourse to their supplemental triangles, and say--The sides of these last triangles cannot but be equal each to each, since they are the respective supplements of equal angles. But this equality of the three sides supposes that of the three angles. The two supplemental triangles are therefore perfectly equal ; consequently their angles have equal supplements. But these supplements are the very sides of the proposed triangles. Therefore iwo spherical triangles are equal, when iheir angles are respectively equal.

This is a remarkable property of spherical triangles; it does not take place, as we already know, in rectilinear triangles; as with respect to these last we can only infer their similitude.

The equality between two spherical triangles also takes place when in each triangle two sides respectively equal, form an equal angle; and also when two angles of the one, equal to two angles of the other, are formed on an equal side.

It is easy to demonstrate these propositions in the same manner as in plane triangles.

72. Let there now be the spherical triangle CAB, whose sides CA and CB I suppose equal, I desire to know whether the angles opposite these sides are also equal.

I take CD-CE, and describe the arcs BD and AE, which gives the two triangles CAE, CBD, perfectly equal. The arc AE is therefore equal to the arc BD: from this I infer the equality of the other two triangles, viz. ABĖ and ABD. Consequently the angle A and the angle B A

B are equal, and hence in every spherical triangle, equal sides are opposite to equal angles.

The inverse proposition may be demonstrated by the supplemental triangle.

73. If we wish to prove, that in a spherical triangle ABC, a greater angle is always opposite to a greater side, we may say

Let A be greater than B; I can describe an arc DA which will make the angle DAB equal to the A4

B angle B. I shall therefore have an isosceles triangle ABD. The are



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