For example, let comp. a be the middle part and the extremes opposite. 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 R2 cos a=R cos b cos c+ sin b sin c cos A. If A 90° cos A=0; hence = R cos a=cos 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 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.). -R cos B cos C, B R2 cos A sin B sin C cos a-R cos B cos C. sin B sin C cos a= 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 tangent 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.), R2 cos B=sin A sin C cos b-R cos A cos C. = a But if A=90°, cot A=0; hence, cot a sin c-cos c cos B or A or Let comp. B be still the middle part and the extremes adja cent. The equation to be applied must then contain the four four parts a, B, c, and A. It is similar to equations (9.). cot a sin =cos c cos B+cot A sin B 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 a=cos b cos c=cot B cot C (10.) 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 b and c will be of the same species, and also that the angles B and C will likewise be of the same 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. LOGARITHMS. SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES BY 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. 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 The hypothenuse a corresponds to the middle part, and the extremes are opposite: hence R cos a=cos b cos c, or ar.-comp. log. As cos Is to To cos b R a C 42° 12' EXAMPLES. 64" 40' 54° 43′ 07′′ 64° 40' 42° 12' To find the angle B. The side b will be the middle part and the extremes opposite: hence R sin b=cos (comp. a) x cos (comp. B)=sin a sin B. As sin a ar.-comp. log. Is to sin b So is R To sin B 48° 00′ 14′′ R As So is tang b 42° 12' To cos C 64° 34′ 46" R cos C=cot a tang b. ar.-comp. A 0.130296 10.000000 9.631326 9.761622 To find the angle C. The angle C is the middle part and the extremes adjacent; hence log. 0.043911 9.827189 10.000000 9.871100 0.000000 9.675237 9.957485 9.632722 2. In 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, As cot Is to a 105° 34' R So is cos B 80° 40′ To tang 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. ar.-comp. log. To find the side c. The angle B will correspond to the middle part, and the extremes will be adjacent: hence, R cos B-cot a tang c. ar.-comp. c 149° 47' 36" 0.784220+ 9.428717 10.000000+ 10.212937 log. To find the side b. The side b will be the middle part and the extremes opposite: hence, 0.55505310.000000 + 9.209992+ 9.765045 a 0.000000 9.983770 9.994212 9.977982 OF QUADRANTAL TRIANGLES. 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°. If we pass 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 we see, that the polar triangle will be A right angled at A', and hence every case may be referred to a right angled triangle. But we can solve the quadrantal triangle by means of the right angled triangle in a manner still more simple. In the quadrantal triangle BAC, in which BC 90°, produce the side CA till CD is equal to 90°, and conceive the arc of a great circle to be drawn through B and D. Then C will be the pole of the arc BD, and the angle C will be measured by BBD (Book IX. Prop. VI.), and the angles CBD and D will be right angles. Now before the remaining parts of the quadrantal triangle can be found, at least two parts must be given in addition to the side BC=90°; in which case two parts of the right angled triangle BDA, together with the right angle, become known. Hence the conditions which enable us to determine one of these triangles, will enable us also to determine the other. 3. In the quadrantal triangle BCA, there are given CB=90°, the angle C-42° 12', and the angle A=115° 20' required the remaining parts. As sin To sin d = A Having produced CA to D, making CD=90° and drawn the arc BD, there will then be given in the right angled triangle BAD, the side a=C=42° 12', and the angle BAD=180°— BAC=180°-115° 20' 64° 40', to find the remaining parts. R sin a sin A sin d. = ar.-comp. To find the side d. The side a will be the middle part, and the extremes opposite: hence, 64° 40' 42° 12' 48° 00′ 14′′ a As cos a 42° 12' to R So is cos A To sin B d 64° 40' 40 35° 16′ 53′′ ΤΑ ib log. To find the angle B. The angle A will correspond to the middle part, and the extremes will be opposite : hence R cos A sin B cos a. = ar.-comp. log. 0.043911 10.000000 9.827189 9.871100 0.130296 10.000000 9.631326 9.761622 |