First, when A is acute and a less than b. Calculate y from cos y = cos a + cos and V: and a': from tan : cot b's = Cos b LID A, cos a': = There are two triangles which satisfy the data ; in the first, c = x — y, n' is the angle opposite to b, and c = 6"2 ; in the second, c = x + y, B is the angle opposite to it and c = L'z + z.
Secondly, when A ie acute and a equal to or greater than 6. Cal culate exactly as in the last case, but there is only one triangle which satisfies the data, namely, the second of the preceding.
Thirdly, when A is obtuse, in which case there is no triangle, unless a be greater than b. Calculate (using A as more convenient) sin it ein = sin 6 sin A', tan x = tau b cos A, sin B = sin A' 8a Use the value of B which is less than a right angle ; calculate b": and a': from tan z cot Is'a= cos 6 tau A', cos a': a' Then c y — x, and c = a': — tez.
lu the case in which one or both of the sides are greater than a right angle, which rarely, if ever, occurs, it is beat to have recourse to one of the adjacent triangles described at the beginning of this article, and to use it in the same manner as the supplementary triangle has been used. It is not however necessary to dwell on this point.
Supplement.—Given two angles (A and n) and a side opposite to one of them (a); required the remaining parts. Let A' and a be the sides of a triangle, and a' the angle opposite to A'. Find c' the remaining side, and 6' and c' the remaining angles ; then c is the remaining angle of the original triangle, and it and c the remaining sides.
All the cases would need some subdivision to adapt them to cal culation, if it were really often required to solve triangles with very largo sides and angles. But in application it generally happens that the reasoning of the previous part of the process is so conducted as to throw the calculation entirely upon triangles which have at least two sides and two angles severally less than a right angle. Divide a great circle into three parts, and we have the extreme limit of a spherical triangle : the sum of its sides being 360', and the sum of its angles six right angles. But a triangle which should be very near to this limit would be best used in reasoning, and solved in practice, by means of one of the other seven triangles into which its circles divide the sphere. And if one of the sides should be greater than two right
angles, the remainder of the hemisphere would be the triangle on which calculation is employed. And it is to be understood that all the formulas are demonstrated only for the case in which every side is less than two right angles. At the same time we should recommend the beginner to procure a small sphere, and to habituate himself to the appearance of all species of triangles.
The area of a spherical triangle is singularly connected with the sum of its angles, on which alone it depends, the sphere being given. Let any two triangles, however differently formed, have the sum of their angles the same, and they must have the same area. If the angles be measured in theoretical unite [AaOLE], the formula is as follows : r being the radius, Area = (A + + r), which gives the number of square units in the area, the radius being expressed in corresponding linear units. But if the angles be measured in degrees and fractions of a degree, the formula is Area = .01745329252 (A + n + c — 180).
The angle A + a + c 180° is called the spherical excess, and it may be found at once from the sides by the formula eph ex. a a—a a--b a—c tan= 2 2 2 2 =tan — tan -- tan — tan • --2 ' so that the area of a triangle is easily found from its aides. If a spherical triangle were flattened into a plane one, without any altera tion of the lengths of its silos, it is obvious that the sum of the angles would undergo a diminution, being reduced to 180°. The angles would not diminish equally; but, if the sphericity of the original triangle were small, or If it occupied only a small part of the sphere, the diminutions which the several angles would undergo in the process of being flattened would be so nearly equal, that it would be useless, for any practical purpose, to consider them M unequal. For a triangle of HMO sphericity, then, it may be assumed that in being flattened, each of its angles loses one-third of the spherical excess. This pro position is one of considerable use hi the measurement of a degree of the meridian.