Imágenes de páginas
PDF
EPUB

CHAPTER XVIII

RELATIONS AMONG THE SIDES AND THE ANGLES OF A
SPHERICAL TRIANGLE

174. The succeeding pages contain a brief discussion of some of the properties of spherical triangles.

For the sake of ready reference, we shall enumerate, without proof, some properties of solid figures. The statements contained in sections 175-177 are proved in works on solid geometry, to which the student is referred.

The curve of intersection of a

175. Definitions and Theorems. plane and a sphere is a circle.

When the plane of the circle passes through the centre of the sphere, their curve of intersection is called a great circle.

One great circle can be passed through any two points on the surface of the sphere, and only one if these points are not extremities of a diameter of the sphere.

A spherical figure is any part of the surface of the sphere bounded by arcs of great circles.

A spherical polygon is a spherical figure bounded by more than two arcs. The arcs are called the sides of the polygon. The intersections of the arcs are called the vertices of the polygon.

176. The angle between two great circles is measured by the angle between the tangents drawn to the circles at their point of intersection. This is called a spherical angle.

The angle between two great circles equals the angle between their planes.

177. SPHERICAL TRIANGLES. A spherical triangle is a spherical polygon of three sides.

Let ABC be a spherical triangle.

Let O be the centre of the sphere.

By the letters A, B, C we shall indicate geometrically the three angular points of the triangle ABC; algebraically, the three angles at those points respectively. By the letters a, b, c we shall indicate the measures of the sides opposite A, B, C, respectively.

A

B

FIG. 63.

a, b, c are measured by the angles at the centre of the sphere, and hence they are measured in angular units; e.g. c is measured by angle AOB.

We know, then, the following properties:

=

--

A 180° — a';
B = 180° — b';
C = 180° - c';

a

1. *The sum of two sides of a triangle is greater than the third. 2. The greatest side is opposite the greatest angle, and conversely. 3. Any angle A<180°.

4. (A+B+ C) < 540° and > 180°.

5. Any side a < 180°.

6. (a+b+c) < 360°.

7. AB and a b are of the same sign.

8. A side which differs from 90° more than another side is in the same quadrant as the angle opposite it.

9. If A'B'C' is the polar triangle of ABC, and if A', B', C' are its angles, and a', b', c' the corresponding sides, then

[blocks in formation]

*NOTE. We know that in general three great circles intersect in such a way as to form eight spherical triangles; and that at least one of these triangles satisfies the conditions of Art. 177. We shall consider only such triangles.

or

i.e.

A

M

[ocr errors]

N

FIG. 64.

178. To prove cos a = cos c cos b + sin b sin c cos A.

Let ABC be a spherical triangle.

Let O be the centre of the sphere, and OA OB OC = radius of the sphere.

In the plane AOB, draw LM perpendicular to OA.
In the plane AOC, draw LN perpendicular to OA.
Then the plane angle MLN = A (Art. 176).
In the plane triangles LMN and MON,

MN=LM2 + LN-2 LM LN cos A.
MN2 = 0M2 + ON2 - 2 OM ON cos a.

α

=

Equate these values of MN',

OM2 – LM2 + ON2 – LN2 - 2 OM ON cos a +2 LM. LN cos A=0. OM - LM = OL, and ON- LN2 = OL";

But

.. 2 OL-2 OM ON cos a +2 LM. LN cos A = 0;

OL OL LM LN
+
OM ON OM ON

cos c cos b+ sin b sin c cos A = cos a.

=

cos A= cos a;

(Art. 107.)

(a)

179. By reference to Fig. 63 it is evident that the demonstration of the preceding article requires that c and b are each less than 90°, while a is unrestricted.

Suppose, now, that one of the sides, b, say, is greater than 90°. Produce the great circles AC and BC. They intersect again in C', Fig. 65.

But

and

and

[blocks in formation]

cos a':

Whence

AB < 90° (hypothesis).

AC' = (180° — AC) < 90°.

We can now apply the formula (a), of the preceding article, to ABC'.

Hence

FIG. 65.

с

= cos b' cos c + sin b' sin e cos (180° — A).
cos a'
―― cos a;
cos b'

cos b;

cos (180° - A) Substituting these values in (b) we obtain

cos A.

[ocr errors]

cos A:

[ocr errors]

cos a = cos b cos c + sin b sin c cos A.

cos B:

cos C

B

Similarly, if both b and c are greater than 90°, it may be shown cos a = cos b cos c + sin b sin c cos A.

a

that

So that (a) is true for all spherical triangles which we are considering. Similarly we can express the other angles in terms of the sides. We therefore have this relation involving the sides and one angle.

=

cos a = cos b cos c + sin b sin c cos A,
cos b = cos c cos a + sin c sin a cos B,
cos c = cos a cos b + sin a sin b cos C.

=

с

[blocks in formation]

(b)

(Art. 59.) (Art. 59.)

COS C cos a cos b

sin a sin b

(1)

(2)

The last two formulæ of either set can be derived from the first one of that set by making a cyclical interchange of a into b, b into c, c into a, and at the same time changing A into B, B into C, C into A.

180. To show that

sin A

sin B

sin C

sin a

sin b

sin c

This is a relation involving the sides and the opposite angles.

PROOF. Sin2 A= 1 - cos2 A

sin' b sin2 c

[blocks in formation]

sin A

sin a

(Art. 179.)

Substitute for sin b, 1 - cos2 b, and sin2 c = 1-cos2 c, and reducing we obtain

sin2 A =

=

Multiplying both sides of the equation by square root of each side, we obtain

=

(cos a sin2b sin2 c

sin B
sin b

=

1-cos2bcos2c- cos2 a + 2 cos a cos b cos c
sin2 b sin2 c

cos b cos c)2

sin C
sin c

sin B
sin b

But this is equivalent

In a similar way we might solve for
to a cyclical interchange of the letters as described in Art. 179. But
an interchange of the letters changes the left side of the equation
into

sin B
sin b'

and leaves the right side unchanged. Hence

√1-cos2bcos2 c cos'a + 2 cos a cos b cos c
sin a sin b sin c

1

sin' a'

and extracting the

(3)

V1-cos2b - cos2 c cos2 a + 2 cos a cos b cos c
sin a sin b sin c

181. We shall add a geometrical proof of the theorem.

Take L any point in OA. plane COB, piercing it in G;

Q.E.D.

since A180°, and since a < 180° (Art. 177).

Therefore sin A and sin a are positive. Therefore there is no ambiguity of sign.

From L let a perpendicular fall on from G draw GM and GN perpen

« AnteriorContinuar »