С Cornp. 16 Comp. a Cornp. In every right angled spherical triangle BAC, there are six parts: three sides and three angles. If we omit the consideration of the right angle, which is always known, there will be five remain в. ing parts, two of which must A be given before the others can be determined. The circular parts, as they are called, are the two sidesc and b, about the right angle, the complements of the oblique angles B and C, and the complement of the hypothenuse a. Hence there are five circular parts. The right angle A not being a circular part, is supposed not to separate the circular parts c and b, so that these parts are considered as adjacent to each other. If any two parts of the triangle be given, their corresponding circular parts will also be known, and these together with a required part, will make three parts under consideration. Now, these three parts will all lie together, or one of them will be separated from both of the others. For example, if B and c were given, and a required, the three parts considered would lie together. But if B and C were given, and b required, the parts would not lie together; for, B would be separated from C by the part a, and from b by the part c. In either case B is the middle part. Hence, when there are three of the circular parts under consideration, the middle part is that one of them to which both of the others are adjacent, or from which both of them are separated. In the former case the parts are said to be adjacent, and in the latter case the parts are said to be opposite. This being premised, we are now to prove the following rules for the solution of right angled spherical triangles, which it must be remembered apply to the circular parts, as already defined. 1st. Radius into the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part is equal to the rectangle of the cosines of the opposile parts. These rules are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, then adjacent. Having thus fixed the three parts which are to be considered, take that one of the general equations for oblique angled triangles, which shall contain the three corresponding parts of the triangle, together with the right angle: then make A=90', and after making the reductions corresponding to this supposition, the resulting equation will prove the rule for that particular case. or For examiple, let comp. a be the middle part and the extremes oppusite. The equation to be applied in this case must contain a, b, c, and A. The first of equations (2.) contains these four quantities: hence Ro cos a=R cos b cos c+sin b sin c cos A. If A=96° cos A=0; hence Rcos arcos b cos c; that is, radius into the sine of the middle part, (which is the complement of a,) is equal to the rectangle of the cosines of the opposite parts. Suppose now that the complement С of a were the middle part and the extremes adjacent. The equation to be applied must contain the four quantities a, B, C, and A. It is the first of equations (8.). B 'A Rocos A=sin B sin C cos Q-R cos B cos C. Making A=90°, we have sin B sin C cos a=R cos B cos C, R cos a=cot B cot C; that is, radius into the sine of the middle part is equal to the rectangle of the tangent of the complement of B into the fangent of the complement of C, that is, to the rectangle of the tangents of the adjacent circular parts. Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under consideration will then be the perpendicular b and the angle C. The equation to be applied must contain the four parts A, B, C, and b: it is the second of equations (8.), Rcos B=sin A sin C cos b-R cos A cos C. Making A=90°, we have, after dividing by R, R cos B=sin C cos b. Let comp. B be still the middle part and the extremes adja: cent. The equation to be applied must then contain the fou: four parts a, B, c, and A. It is similar to equations (9.). cot a sin e=cos c cos B+cot A sin B But if A=90°, cot A=0; hence, cot a sin c=cos c cos B; o And by pursuing the same method of demonstration when each circular part is made the middle part, we obtain the five following equations, which embrace all the cases. R cos arcos 6 cos c=cot B cot C (10.) R sin c=sin a sin C=tang b cot B) We see from these equations that, if the middle part is required we must begin the proportion with radius ; and when one of the extremes is required we must begin the proportion with the other extreme. We also conclude, from the first of the equations, that when the hypothenuse is less than 90°, the sides band will be of the same species, and also that the angles B and C will likewise be of the sáme species. When a is greater than 90°, the sides b and c will be of different species, and the same will be true of the angles B and C. We also see from the two last equations that a side and its opposite angle will always be of the same species. These properties are proved by considering the algebraic signs which have been attributed to the trigonometrical lines, and by remembering that the two members of an equation must always have the same algebraic sign. SOLUTION OF RIGIIT ANGLED SPHERICAL TRIANGLES BY LOGARITHMS. It is to be observed, that when any element is discovered in the form of its sine only, there may be two values for this element, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the element in question is less or greater than 90°; the element will be less than 90°, if its cosine, tangent, or cotangent, has the sign + ; it will be greater if one of these quantities has the sign In order to discover the species of the required element of the triangle, we shall annex the minus sign to the logarithms of all the elements whose cosines, tangents, or cotangents, are negative. Then by recollecting that the product of the two . extremes has the same sign as that of the means, we can at once determine the sign which is to be given to the required element, and then its species will be known. EXAMPLES. с 1. In the right angled spherical triangle BAC, right angled at A, there are given a=64° 40' and b=42° 12': required the remaining parts. First, to find the side c. B 'A The hypothenuse a corresponds to the middle part, and the extremes are opposite : hence R cos a=cos b cos C, As cos 6 42° 12' ar.-comp. 0.130296 Is to R 10.000000 So is cos 64" 40 9.631326 To cos 54° 43'07" 9.761622 or log. a с To find the angle B. The side b will be the middle part and the extremes oppo. site : hence R sin b=cos (comp. a) x cos (comp. B)=sin a sin B. As sin 64° 40' ar.-comp. log. 0.043911 Is to sin b 42° 12' 9.827189 So is R 10.000000 To sin B 48° 00' 14" 9.871100 a To find the angle C. The angle C is the middle part and the extremes adjacent jence R cos C=cot a tang b. ar.-comp. log. 0.000000 Is to cot 64° 40 9.675237 So is tang b 42° 12' 9.957485 C 64° 34' 46". 9.632722 a To cos 2. la a right angled triangle BAC, there are given the hy. pothenuse a=105° 34', and the angle B=80° 40': required the remaining parts. To find the angle C. The hypothenuse will be the middle part and the extremes adjacent: hence, R cos arcot B cot C. As cot B: 80° 40' ar.-comp. log. 0.784220+ Is to cos a 105° 34' 9.428717So is R 10.000000+ To cot C 148° 30'54" 10.212937 Since the cotangent of C is negative the angle C is greater than 90°, and is the supplement of the arc which would correspond to the cotangent, if it were positive. To find the side c. The angle B will correspond to the middle part, and the extremes will be adjacent: hence, Rcos B=cot a tang c. As cot a 105° 34' ar.-comp. log. 0.555053— Is to R 10.000000 + So is cos B 80° 40' 9.209992 + C 149° 47' 36" 9.765045 To tang To find the side b. The side b will be the middle part and the extremes oppo site: hence, R sin b=sin a sin B. ar. comp. 0.000000 To sin a 105° 34' 9.983770 So is sin B 80° 40' 9.994212 To sin 6 71°54' 33" 9.977982 OF QUADRANTAL TRIANGLES. If we pass A quadrantal spherical triangle is one which has one of its sides equal to 90°. Let BAC be a quadrantal triangle in which the side a=90°. to the corresponding polar triangle, we shall have A'=180°-a=90°, B’= 180°—b, C'=180°C, a=180°-A, b'=180°-B,C=180°C; from which B 'A we see, that the polar triangle will be |