« PreviousContinue »
Rules for determining whether the sides and angles of spheri-
when a leg and the adjacent angle are known (585), (586),
1. Spherical Trigonometry treats of the solution of spherical triangles.
A Spherical Triangle is a portion of the surface of a sphere included between three arcs of great circles. In the present treatise those spherical triangles only are treated of, in which the sides and angles are less · than 180°.
2. The angle, formed by two sides of a spherical 2 triangle, is the same as the angle formed by their planes.
3. An isosceles spherical triangle is one which has two of its sides equal.
An equilateral spherical triangle is one which has all its sides equal.
4. A spherical right triangle is one which has a right angle, all other spherical triangles are called oblique. We shall in spherical trigonometry, as we did in
plane trigonometry, attend first to the solution of
Spherical Right Triangles.
Napier's Rules for the Solution of Spherical Right Triangles.
To investigate some relations between the sides and angles of a spherical right triangle.
Solution. The importance of this problem is obvious; for, unless some relations were known between the sides and the angles, they could not be determined from each other, and there could be no such thing as the solution of a spherical triangle.
Let, then, ABC (fig. 1.) be a spherical right triangle, right angled at C. Call the hypothenuse AB, h; and call the legs BC and AC, opposite the angles A and B, p. respectively a and b.
Let O be the centre of the
sphere. Join OA, OB, OC.
The angle A is, by art. 2, equal to the angle of the planes BOA and COA. The angle B is equal to the angle of the planes BOC and BOA. The angle
of the planes BOC and AOC is equal to the angle (428) C, that is, to a right angle; these two planes are, therefore, perpendicular to each other.
Moreover, the angle BOA, measured by BA, is equal to BA or h; BOC is equal to its measure BC (429) or a, and AOC is equal to its measure AC or b.
Through any point A' of the line OA, suppose a plane to pass perpendicular to OA. Its intersections A'C' and A'B', with the planes COA and BOA must (430) be perpendicular to OA', because they are drawn through the foot of this perpendicular.
As the plane B'A'C' is perpendicular to OA, it must be perpendicular to AOC; and its intersection B'C', with the plane BOC, which is also perpendicular to AOC, must likewise be perpendicular to AOC. Hence B'C' must be perpendicular to A'C' and OC' which pass through its foot in the plane AOC.
All the triangles A'OB', A'OC', B'OC', and (431) A'B'C' are then right-angled; and the comparison of them leads to the desired equations, as follows:
First. We have from triangle A'OB' by (5) and