CASE 6.

Let A' and A" be given.

cos Scot A' cot A" (232. IV.)

(257.) COR. 1. Hence, any proposed isosceles spherical triangle may be solved, if any three of its six parts be given.

For, its solution is reduced to that of a right-angled triangle, by drawing, from the vertex, an arch of a great circle (Art. 70.) perpendicular to the base: because (Art. 107.) the angles at the base are equal, and (Art. 109.) the arch, which is perpendicular to the base, bisects the base and the vertical angle.

(258.) COR. 2. If (S) be the given side of an equilateral spherical triangle, and (4) any one of its equal angles (Art. 108.) the angle A may be found by means of Art. 232. VIII. a perpendicular arch having been first let fall, from either of the other angles, upon the opposite side. For, then, since that side is bisected (Art. 109.) by the perpendicular arch, cos A=tan S cot S. 14

(259.) COR. 3. If two of the sides, or two of the angles, of a spherical triangle, be the supplements, each of the other, it is evident, from Art. 210, that the solution of such a triangle may be reduced to that of an isosceles

spherical triangle; which latter solution, as hath been shewn, (Art. 257.) may be made to depend on the solution of a right-angled spherical triangle.

5.4.3

(260.) SCHOLIUM. The five parts exclusive of the right angle, of a right-angled spherical triangle, admit of ten combinations, when they are taken three and three together: but, of these, only six are essentially different from each other. Now, every one of these six different combinations comprehends three problems; for any two of the quantities, out of each set of three, may be the quantities that are given. There are, therefore, in the whole, eighteen such problems. They cannot, all of them, however, be considered as really different problems; because, in some of the combinations, two of the three problems, when they are stated in general terms, become, in effect, one and the same problem.

Again, the parts, which are sought, are found in terms of their trigonometrical functions. Whenever, therefore, any required part is expressed in terms of its sine, the part itself (Introd. 14. 19.) is only so far determined, as to be known to be one of two different arches, or angles, which are supplements of each other: so that, in reality, there are then two different triangles which satisfy the general conditions of the problem. This is the case, when either the base, or the perpendicular, and the opposite angle, are given, to find (V. VI. VII.) the other parts. There remains, therefore, in this case, an am

biguity; unless it can be removed, by the application of Art. 127. 129.

But, in all the other cases, in which the unknown part is expressed in terms of its cosine, of its tangent, or of its co-tangent, the sign of the value of that function will shew whether the part sought be greater, or less, than a quadrant, or than a right angle. It will be less (Introd. 19.) than that quantity, if the value of its cosine, its tangent, or its co-tangent, be positive, and greater, when that value is negative. Here, therefore, the ambiguity, which (Introd. 14.) might otherwise obtain, is removed by the algebraic sign of the function and it may, also, be removed, by the application of Art. 130.

The following Table may be used as a key to the solution of all the cases of right-angled spherical triangles : Any two quantities, in each of the six specified combinations, being given, the third may be found, by the Forms and articles to which a reference is made.

(1.) Hypotenuse, and the other two Sides

(III. and Art. 130.)

(2.) Hypotenuse and two oblique Angles

(IV. and Art. 130.)

(3.) Hypotenuse, one other Side and adjacent Angle (VIII. and Art. 130.)

(4.) Hypotenuse, one other Side and opposite Angle (V. and Art. 127. or else ambiguous.)

(5.) Base, Perpendicular and either oblique Angle

(VII. and Art. 127. or else ambiguous.)

(6.) Two oblique Angles, and Base, or Perpendicular

(VI. and Art. 127. or else ambiguous.)

It may, further, be remarked, that, when a rightangled spherical triangle is proposed, for solution, the Forms which have been investigated, serve, also, to determine whether the problem be possible. The following conclusions may easily be derived from them.

In right-angled spherical triangles, having only one angle a right angle,

1. The complement of the hypotenuse cannot exceed the complement of either of the other sides (III.) 2. The hypotenuse may have any magnitude, less than 180°, relative to that of either of the oblique angles (IV.)

3. The complement of either of the two oblique angles cannot be less than the complement of the opposite side (V. or VI.)

4. Either of the two oblique angles may have any magnitude, less than 180°, with respect to the side adjacent to it (VIII.)

5. But the two oblique angles must be, together, greater than a right angle (VI. or Art. 82.)

6. The two sides, containing the right angle, may have any relative magnitude.

In consequence of the want of absolute exactness, in the logarithmic Tables, which are in use, the Forms that have been investigated for the purpose of solving right

angled spherical triangles, will not yield results sufficiently correct, if the sines and the cosines of the required parts be very great; that is, if they approximate to the value of the radius. In these cases, therefore, it becomes necessary to substitute other equivalent expressions, which may exhibit the parts, that are sought, in terms of their tangents.

PROP. II.

(261.) Problem. The value of the sine, or the cosine, of any required part, of a right-angled spherical triangle, having been found in terms of the given parts, to find the value of the tangent, or of the co-tangent, of the half of that part.

Let A denote the right angle, and S the hypotenuse; A' one of the other angles, and S' the side opposite to it; A" the third angle, and S" the third side, of a rightangled spherical triangle; and let unity be put for the radius: then, (Art. 232. IV.)