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 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.). B R2 cos A-sin B sin C cos a-R cos B cos C. Making A 90°, we have sin B sin C cos a=R cos B cos C, or R cos a=cot B cot C; 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. Making A 90°, we have, after dividing by R, Let R cos B-sin C cos b. comp. B be still the middle part and the extremes adjacent. The equation to be applied must then contain the four four parts a, B, c, and A. It is similar to equations (7.). cot a sin c=cos c cos B+ cot A sin B. But if A=90°, cot A=0; hence, cot a sin ccos c cos B.; or, R cos B-cot a tang c. 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 ccot 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 band 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. 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. The hypothenuse a corresponds to the middle part, and the extremes are adjacent; hence 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. The angle C is the middle part and the extremes adjacent : hence 2. In a right angled triangle BAC, there are given the bypothenuse 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, 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, The side b will be the middle part and the extremes oppo site: hence, A quadrantal spherical triangle is one which has one of its sides equal to 90o.. 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 a d A 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. 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. = To find the side d. The side a will be the middle part, and the extremes opposite: hence, The angle A will correspond to the middle part, and the ex tremes will be opposite: hence |