IN right angled spherical triangles, the co-fine of an angle is to the radius, as the tangent of the fide adjacent to that angle is to the tangent of the hypotenuse. The same construction remaining : In the triangle CEF, sin EF : R:: tan CE : tan CFĚ (18.); but sin EF = cos ABC; tan CE = cot BC, and tan CFE = cot AB, there. fore cof ABC:R:: cot BC : cot AB. Now, because (Cor. def. Pl. Tr.) cot BC :R::R:tan BC, and cot AB : R::R:tan AB, by equality inversely, cot BC : cot AB :: tan AB : tan BC; therefore (u. 5.) cof ABC:R::tan AB : tan BC. Therefore, &c. Q. E. D. I. 9. From the demonstration it is manifeft, that the tangents of any two arches AB, BC are reciprocally proportional to their co-tangents. COR. 2. Because cof ABC:R::tan AB : tan BC, and R:cot BC :: tan BG :R, by equality, cof ABC : cot BC:: tan AB: R. That is, the co-fine of any of the oblique angles is to the cotangent of the hypotenuse, as the tangent of the fide adjacent to the angle is to the radius. PROP. IN N right angled spherical triangles, the co-fine of either of the fides is to the radius, as the cosine of the hypotenuse is to the co-line of the other side. The same construction remaining : In the triangle CEF, fin CF :R::sin CE: fin CFE, (19.); but sin CF =cof CA, fin CE = cos BC, and sin CFE cof AB; therefore, cof CA:R:: cof BC ; cof AB. C.E.D. N right angled spherical triangles, the co-line of either of the sides is to the radius, as the co-fine of the angle opposite to that fide is to the fine of the other angle. I The same construction remaining : In the triangle CEF, fin CF:R::fin EF : fin ECF, (19.); but fin CF=cos CA, fin EF =cof ABC, and fin ECE fin BCA; therefore, cos CA :R :: cof ABC : fin BCA, Q. E. D. IN lique angled, the fines of the sides are proportional to the fines of the angles opposite to them. First, Let ABC be a right angled triangle, having a right angle at A ; therefore, (19.) the line of the hypotenuse BC is to the radius, (or the fine of the right angle at A), as the fine of the fide AC to the fine of the angle B. C And, in like manner, the fine of BC is to the fine of the angle A, as the fine of AB to the fine of the angle C; wherefore (U. 5.) the fine of the side AC is to the fine of the angle B, as the Α. fine of AB to the line of the angle C. Secondly, 'Let ABC be an oblique angled triangle, the fine of any of the sides BC, will be to the fine of any of the other two AC, as the fine of the angle A oppofite to BC, is to the fine of the angle B opposite to AC. Through the point C, let there be drawn an arch of a great circle CD perpendicular to AB; and in the right angled triangle BCD, fin BC:R :: fin CD : fin B, (19.); and in the triangle ADC, fin AC:R :: fin CD:fin A ; wherefore, by equality inversely, sin BC : fin AC :: sin A : fin B. In the Tame manner, it will be proved that sin BC: fin AB :: fin A :sin C, &c. Therefore, &c. Q.E.D. PROP. IN lar arch being drawn from any of the angles upon the opposite side, the co-fines of the angles at the base are proportional to the fines of the legments of the vertical angle. Let ABG be a triangle, and the arch CD perpendicular to the base BA; the co-line of the angle B will be to the co-fine of the angle A, as the fine of the angle BCD to the fine of the angle ACD. For having drawn CD perpendicular to AB, in the right angled triangle BCD, (23.) cof CD:R :: cof B : sin DCB ; and in the right angled triangle ACD, cof CD:R :: cof A: fin ACD; therefore (11. 5.) cof B : fin DCB : : cof A : fin ACD, and alternately, cof B : cof A::fin BCD:fin ACD. Q. E. D. 'HE same things remaining, the co-fines of the fides BC, CA, are proportional to the co-fines of BD, DA, the segments of the base. For in the triangle BCD, (22.), cof BC : cof BD:: cof DC : R, and in the triangle ACD, cos AC: cos AD: : col DC :R; therefore (11. 5.) cof BC : cof BD :: cof AC: cof AD, and alternately, cof BC: cof AC::cos BD: cos AD. Q. E. D. THE same construction remaining, the fines of DA, the segments of the base are reciprocally proportional to the tangents of B and A, the angles at the base. T BD In the triangle BCD, (18.) fin BD:R::tan DC : tan B; and in the triangle ACD, fin AD:R:: tan DC : tan A; therefore, by equality inversely, fin BD : sin AD : ; tan A : tan B. Q. E. D. THE fame construction remaining, the co-fines of the segments of the vertical angle are reciprocally proportional to the tangents of the sides. Because (21.) cof BCD:R:: tan CD : tan BC, and also, cos ACD: R:: tan CD : tan AC, by equality inversely, cos BCD : cof ACD : ; tan AC : tan BC. Q. E. D. PROP. |