this shews that (AB) is positive, negative, or zero, according as (ab) is positive, negative, or zero; thus we obtain all the results included in Arts. 33...36.

57. If two triangles have two sides of the one equal to two sides of the other, each to each, and likewise the included angles equal, then their other angles will be equal, each to each, and likewise their bases will be equal.

We may shew that the bases are equal by applying the first formula in Art. 39 to each triangle, supposing b, c, and A the same in the two triangles; then the remaining two formulæ of Art. 39 will shew that B and C are the same in the two triangles.

It should be observed that the two triangles in this case are not necessarily such that one may be made to coincide with the other by superposition. The sides of one may be equal to those of the other, each to each, but in a reverse order, as in the following figures.

Two triangles which are equal in this manner are said to be symmetrically equal; when they are equal so as to admit of superposition they are said to be absolutely equal.

58. If two spherical triangles have two sides of the one equal to two sides of the other, each to each, but the angle which is contained by the two sides of the one greater than the angle which is contained by the two sides which are equal to them of the other, the base of that

which has the greater angle will be greater than the base of the other; and conversely.

Let b and c denote the sides which are equal in the two triangles; let a be the base and A the opposite angle of one triangle, and a' and A' similar quantities for the other. Then

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

cos a' = cos b cos c + sin b sin c cos A';

sin † (a + a') sin } (a – a') = sin b sin c sin 1 (A + A') sin § (A – A');

this shews that

59. If on a sphere any point be taken within a circle which is not its pole, of all the arcs which can be drawn from that point to the circumference, the greatest is that in which the pole is, and the other part of that produced is the least; and of any others, that which is nearer to the greatest is always greater than one more remote; and from the same point to the circumference there can be drawn only two arcs which are equal to each other, and these make equal angles with the shortest arc on opposite sides of it.

This follows readily from the preceding three Articles.

60. We will give another proof of the fundamental formulæ in Art. 39, which is very simple, requiring only a knowledge of the elements of Co-ordinate Geometry.

Suppose ABC any spherical triangle, O the centre of the sphere, take O as the origin of co-ordinates, and let the axis of z pass through C. Let x, y, z, be the co-ordinates of A, and x, y, those of B; let r be the radius of the sphere. Then the square on the straight line AB is equal to

2

and x2+ y2+z,2 = r2, x, ̧2+Y22+222 = r2, thus

Now make the usual substitutions in passing from rectangular to polar co-ordinates, namely,

cos e, cos 0, + sin 0, sin 0, cos (☀, — 4,) = cos AOB,

2

that is, in the ordinary notation of Spherical Trigonometry,

cos a cos b + sin a sin b cos C = COS C.

This method has the advantage of giving a perfectly general proof, as all the equations used are universally true.

EXAMPLES.

1. If Aa, shew that B and b are equal or supplemental, as also C and c.

2. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle.

3. When does the polar triangle coincide with the primitive triangle?

4. If D be the middle point of AB, shew that

cos AC + cos BC = 2 cos AB cos CD.

5. If two angles of a spherical triangle be respectively equal to the sides opposite to them, shew that the remaining side is the supplement of the remaining angle; or else that the triangle has two quadrants and two right angles, and then the remaining side is equal to the remaining angle.

α

In an equilateral triangle, shew that 2 cos sin

α

2

7. In an equilateral triangle, shew that tan2 = 1-2 cos 4; hence deduce the limits between which the sides and the angles of an equilateral triangle are restricted.

8. In an equilateral triangle, shew that sec A = 1 + sec a.

9. If the three sides of a spherical triangle be halved and a new triangle formed, the angle

b

b

between the new sides and 02

с

is given by cos = cos A + 1⁄2 tan tan sin2 0.

2

2

10. AB, CD are quadrants on the surface of a sphere intersecting at E, the extremities being joined by great circles: shew that

COS AEC: cos AC cos BD - cos BC cos AD.

11. If b + c = π, shew that sin 2B + sin 2C = 0.

12. If DE be an arc of a great circle bisecting the sides AB, AC of a spherical triangle at D and E, P a pole of DE, and PB, PD, PE, PC be joined by arcs of great circles, shew that the angle BPC twice the angle DPE.

13. In a spherical triangle shew that

sin b sin c + cos b cos c cos Asin B sin C-cos B cos C cos a.

14. If D be any point in the side BC of a triangle, shew that cos AD sin BC= cos AB sin DC + cos AC sin BD.

15. In a spherical triangle shew that if 0, 4, be the lengths of arcs of great circles drawn from A, B, C perpendicular to the opposite sides,

= √(1 - cos2 a - cos3b - cos3 c + 2 cos a cos b cos c).

16. In a spherical triangle, if 0, o, be the arcs bisecting the angles A, B, C respectively and terminated by the opposite sides,

17. Two ports are in the same parallel of latitude, their common latitude being 7 and their difference of longitude 2λ: shew that the saving of distance in sailing from one to the other on the great circle, instead of sailing due East or West, is

2r {λ cos - sin1 (sin à cos 7)},

A being expressed in circular measure, and r being the radius of the Earth.

18. If a ship be proceeding uniformly along a great circle and the observed latitudes be 1,,, at equal intervals of time, in each of which the distance traversed is s, shew that

r denoting the Earth's radius: and shew that the change of longitude may also be found in terms of the three latitudes.

V. SOLUTION OF RIGHT-ANGLED TRIANGLES.

61. In every spherical triangle there are six elements, namely, the three sides and the three angles, besides the radius of the sphere, which is supposed constant. The solution of spherical triangles is the process by which, when the values of a sufficient number of the six elements are given, we calculate the values of the remaining elements. It will appear, as we proceed, that when the values of three of the elements are given, those of the remaining three can generally be found. We begin with the right-angled triangle: here two elements, in addition to the right angle, will be supposed known.