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

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

H This premised, prolong the two arcs CA, CB, till they meet the arc DE: we shall have EG-909, 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+AI, 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 seoond triangle will be reciprocally the supplements 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 reciprecally 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.

2y. 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 two 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 trian

D gles 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 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

C 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 А?

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

Сс

A

B

BCD will then be equal to AD+DC. But AD+DC is visibly greater than AC; therefore the arc BC, opposite to the angle A, is greater than the arc AC, opposite to the angle B.

With the assistance of the supplemental triangle, the inverse proposition offers no difficulty. 74. In the spherical triangle ABC,

D

E supposed to be right-angled at A, it may happen that the angle B is opposite to an arc less than 90°, as AC; exactly 90°, as AD; or greater than 90°, as the arc AE. Of what species in any one of these AS cases, will be the angle at B?. Will it

B be acute, right, or obtuse ?

Since the arc AD is 90°, and is besides perpendicular to the arc AB, the point D is necessarily the pole of AB; therefore from this point, letting fall the arc DB, we shall have a right angle DBA. Hence the angle CBA will be acute, and the angle EBA will be obtuse. Hence the angle B is of the same species as the side opposed to it. The same is the case with respect to the angle C.

75. To distinguish these angles from that which we have supposed a right angle, they are called oblique angles. And we may say that in any right-angled spherical triangle, each of the oblique angles is of the same species as the side opposite to il.

This denomination of oblique angles does not exclude them from being right, as well as the angle A. They may even be obtuse ; but to avoid circumlocution, they are all called oblique. The hypothenuse is the side opposite that right angle which is considered as such. Thus BC is the hypothenuse of the triangle BAC.

76. As the two sides of the right-angle may be of the same or of different species, it is desirable to know before hand of what species the hypothenuse will be in each of these cases.

Suppose then at first that AC and AB are each less than 90°, the angle ACB will be acute, and consequently its supplement BCD will be obtuse. The side BD opposite to this supplement in the triangle DBC, will be greater than the side BC, opposite the angle CDB (73) which ought to be acute for the same reason as the angle AB. But the side BD is only 90° ; therefore the hypothenuse BC must be less than 90'.

Again suppose that AC and AB are each more than 90°, and describe an arc BD, cutting the arc AC, so that AD may be 900

This arc will also be 90°.
But the angle C is obtuse; the angle
ADB is the same; therefore its supple. A
ment BDC is acute, and being so, it must

B be opposite to a side in the triangle BCD, less than that which is opposite to the angle C. Consequently the hypothenuse AC is less than BD; that is to say, in this second case also, as well as in the first, it is less than 900.

D

It remains to shew 'what will be the result on the supposition that one side of the right-angle is greater, and the other less than 90°.

For example, let AB be greater than the quadrant AD; and AC less than AD: we shall have the arc CD, 90°, and the angle CDA will be acute, therefore CDB will be obtuse, and consequently the arc BC, opposite to it, will be greater than

B CD, and therefore more than 90°.

77. From what we have shewn, it follows that in a spherical triangle, if the two sides of the right-angle are of the same species, the hypothenuse will be less than 90°; and if they are of different species, the hypothenuse will be grealer than 90..

And as the oblique angles are always of the same species as the sides opposite to them, they equally serve to shew of what species the hypothenuse is.

78. And reciprocally the hypothenuse may serve to shew of what species are the sides of the right and oblique angles. If, for example, the hypothenuse, and one of the sides be of the same species, the other side is less than 90°; and if the hypothenuse and one of the sides be of different species, the other side is more than 90.

Observe, however, that in the case where one of the sides of the right-angle is 90°, it may happen likewise that the other side is 90°, as well as the first. In this case the three angles are evidently right-angles. It may also happen that this other side may be more or less than 90°. We shall explain the mode of solving spherical triangles in these different cases.

79. In plane trigonometry it has been shewn that the sines of the angles are proportional to the sides opposite these same angles. But as in spherical triangles, the sides are circular arcs, the same proportion cannot obtain here. We shall endeavour to substitute another which applies to all sorts of triangles.

PRINCIPLES AND PROPORTIONS FOR THE SOLUTION OF SPHERICAL TRIANGLES.

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Drawing the line DI, it is evident that the triangle CDI, rightangled at I, is similar to the triangle FGE, right-angled in G; we have therefore

FE: CD::FG : CI that is, radius is to the sine of the hypothenuse, as the sine of the angle B is to the sine of the opposite arc CA. In like manner it may be proved that radius is to the sine of the hypothenuse, as the sine of the angle C is to the sine of the opposite arc AB. Generally, therefore,

81. In every right-angled spherical triangle, radins is to the sine of the hypothenuse, as the sine of one of the oblique angles is to the sine of the side opposite to ii. 82. If the trian

C
gle ABC is oblique-
angled, letting fall
the perpendicular arc
CD, we shall have
the two following

pro-
A

A

B portions ; by (81)

D

B
R: sin AC::sin A : sin CD

R: sin BC::sin B : sin CD
Therefore,

sin A : sin B::sin BC : sin AC and consequently in any spherical triangle, the sines of the angles are proportional to the sines of the sides.

83. Suppose now the triangle ABC to be right-angled at A, and the sides BC and AC

E to be prolonged to 90°, the one in D and the other in F; if we draw the arc DE, we shall have the triangle CDE, right-angled in E, whose four parts CE, CD, DE, and D, are respectively the complements of the four BC, AC, B, and AB, of the triangle ABC, as may easily be shewn. Triangles formed in this manner are called complemental triangles. Now A

B in the complemental triangle CDE, we have

R: sin CD::sin D : sin CE or,

R: cos AC::cos AB : cos BC Therefore in any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the cosine of the other side is to the cosine of the hypothenuse.

And consequently, if an oblique-angled spherical triangle be divided into two righi-angled triangles by an arc perpendicular to its base, we shall always have the cosines of the segments of the base proportional to the cosines of the two adjacent sides.

So that in the triangle ABC, for example, after having described

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