2. Find the angle in time of which the log. tang. is

7. Find the angle in time whose log, secant is 10.23456.

Ans. 3h 37m 26.

8. Find the angle in time whose log. cosecant is 10.12346.

Ans. 3h 15m 15o,

Right and oblique triangles.

8. 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.

9. 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 right triangles.

13*

Investigation of Napier's Rules.

CHAPTER II.

SPHERICAL RIGHT TRIANGLES.

10. Problem. 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. 29) 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, 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 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 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'

Investigation of Napier's Rules.

with the planes COA and BOA must 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 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 (4)

hence, from the equality of the second members of these

equations,

cos. hcos, a cos. b.

(288)

Investigation of Napier's Rules.

Secondly. From triangle A'B'C' we have by (4), and the

and, from triangles A'OC′ and A'OB', by (4),

Thirdly. Corresponding to the preceding equation1 between the hypothenuse h, the angle A, and the adjacent side b, there must be a precisely similar equation between the hypothenuse h, the angle B, and the adjacent side a; which

is

cos. B tang. a cotan. h.

(290)

Fourthly. From triangles B'OC BOA', and B'A'C",