COR. Because cos BA: R:: cot ABC: tan ACB, and (Cor. 1. def. 9. Pl. Tr.) cot ACB: R: : R: tan ACB, ex æquo, cot ACB: cos BC :: R: cot AВС. the PROP. XXI. In right angled spherical triangles, the cosine of an angle is to the radius as tangent of the si side adjacent to that angle is to the tangent of the hypotenuse. The same construction remaining; in the triangle CEF, sin FE: R:: tan CE: tan CFE (18.); but sin EF = cos ABC; tan CE = cot BC, and tan CFE= cot AB, therefore cos ABC: R:: cot BC: cot AB. Now, because (Cor. 1. def. 9. 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 (11. 5.) cos ABC: R:: tan AB: tan BC. Therefore, &c. Q. E. D. Cor. 1. From the demonstration it is manifest, that the tangents of any two arches AB, BC are reciprocally proportional to their cotangents. COR. 2. Because cos ABC: R::tan AB: tan BC, and R: cot BC:: tan BC: R, by equality, cos ABC: cot BC:: tan AB: R. That is, the cosine of any of the oblique angles is to the cotangent of the hypotenuse, as the tangent of the side adjacent to the angle is to the radius. PROP. XXII. In right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the hypotenuse is to the cosine of the other side. The same construction remaining: In the triangle CEF, sin CF: R::sin CE: sin CFE, (19.); but sin CF = cos CA, sin CE = cos BC, and sin CFE=cos AB; therefore, cos CA: R::cos BC: cos AB. Q. E. D. PROP. XΧΙΙΙ. In right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle. The same construction remaining: In the triangle CEF, sin CF: R:: sin EF: sin ECF, (19.); but sin CF = cos CA, sin EF = cos ABC, and sin ECF=sin BCA; therefore, cos CA: R::cos ABC: sin BCA. Q. E. D. PROP. XXIV. In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them. First, Let ABC be a right angled triangle, having a right angle at A; therefore, (19.) the sine of the hypotenuse BC is to the radius, (or the sine of the right angle at A), C as the sine of the side AC to the sine of the angle B. And, in like B Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the sine of the angle A opposite to BC, is to the sine 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 trian gle BCD, sin BC: R::sin CD: sin B, (19.); and in the triangle ADC, sin AC: R:: :: sin CD: sin A; : wherefore, by equality inverse1y, sin BC: sin AC :: sin A : sin B. In the same manner, it may be proved that sin BC : sin AB :: sin A: sin C, &c. Therefore, &c. Q. E. D. : PROP. XXV. In oblique angled spherical triangles, a perpendicular arch being drawn from any of the angles upon the opposite side, the cosines of the angles at the buse are proportional to the sines of the segments of the vertical angle. Let ABC be a triangle, and the arch CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD. For having drawn CD perpendicular to AB, in the right angled triangle BCD, (23.) cos CD: R :: cos B: sin DCB; and in the right angled triangle ACD, cos CD: R::cos A: sin ACD; therefore (11. 5.) cos B: sin DCB :: cos A: sin ACD, and alternately, cos B: cos A :: sin BCD : sin ACD. Q. E. D. PROP. XXVI. The same things remaining, the cosines of the sides BC, CA, are proportional to the cosines of BD, DA, the segments of the base. For in the triangle BCD, (22.), cos BC : cos BD :: cos DC: R, and in the triangle ACD, cos AC: cos AD :: cos DC : R; therefore (11. 5.) cos BC: cos BD :: cos AC: cos AD, and alternately, cos BC: cos AC:: cos BD : cos AD. Q. E. D. PROP. XXVII. The same construction remaining, the sines of BD, DA the segments of the base are reciprocally proportional to the tangents of B and A, the angles at the base. In the triangle BCD, (18.), sin BD : R :: tan DC: tan B; and in the triangle ACD, sin AD : R :: tan DC: tan A; therefore, by equality inversely, sin BD : sin AD :: tan A : tan B. Q. E. D. PROP. XXVIII. The same construction remaining, the cosines of the segments of the vertical angle are reciprocally proportional to the tangents of the sides. Because (21.), cos BCD : R :: tan CD: tan BC, and also cos ACD: R :: tan CD: tan AC, by equality inversely, cos BCD: cos ACD :: tan AC: tan BC. Q. E. D. PROP. XXIX. If from an angle of a spherical triangle there be drawn a perpendicular to the opposite side, or base, the rectangle contained by the tangents of half the sum, and of half the difference of the segments of the base is equal to the rectangle contained by the tangents of half the sum, and of half the difference of the two sides of the triangle. Let ABC be a spherical triangle, and let the arch CD be drawn from the angle C at right angles to the base AB, tan (m+n)xtan (m-n)=tan (a+b)×1⁄2 tan (a-b). Let BC=a, AC=b; DD=m, AD=n. Because (26.) cos a: cos b:: cosm: cos n, (E. 5.) cos a+b: cos a-cos b :: cos m+cosn: cos m-cos n. But (1. Cor. 3. Pl. Trig.), cos a+cosb: cosa cos b :: cot (a+b): tan (a−b), and also, cos m+cosn:cosm-cos n :: cot (m+n): tan (m-n). Therefore, (11.5.) cot (a+b): tan (a−b) :: :: cot (m+n): tan (m-n). And because rectangles of the same altitude are as their bases, tan (a+b) × cot (a+b): tan(a+b)×tan (a-b) :: tan (m+n)×cot (m+n) : tan(mxn)+tan (m-n). Now the first and third terms of this proportion are equal, being each equal to the square of the radius, (1. Cor. Pl. Trig.), therefore the remaining two are equal, (9. 5.) or tan (m+n) × tan (m-n)=tan(a+b) × tan (a−b); that is, tan (BD+AD) × tan (BD-AD) = tan(BC+AC) × tan (BC-AC). Q. E. D. COR. 1. Because the sides of equal rectangles are reciprocally proportional, tan (BD+AD): tan (BC+AC) :: tan (BC-AC): tan (BD-AD). COR. 2. Since, when the perpendicular CD falls within the triangle, BD+ AD=AB, the base; and when CD falls without the triangle BD-AD-AB, therefore in the first case, the proportion in the last corollary becomes tan (AB): tan(BC + AC):: tan (BC-AC): tan (BD-AD); and in the second case, it becomes by inversion and alternation, tan(AB): tan (BC+AC) :: tan (BC-AC): tan (BD+AD). THE preceding proposition, which is very useful in spherical trigonometry, may be easily remembered from its analogy to the proposition in plane trigonometry, that the rectangle under half the sum, and half the difference of the sides of a plane triangle, is equal to the rectangle under half the sum, and half the difference of the segments of the base. See (K. 6.), also 4th Case Pl. Tr. We are indebted to NAPIER for this and the two following theorems, which are so well adapted to calculation by Logarithms, that they must be considered as three of the most valuable propositions in Trigonometry. PROP. XXX. If a perpendicular be drawn from an angle of a spherical triangle to the opposite side or base, the sine of the sum of the angles at the base is to the sine of their difference as the tangent of half the base to the tangent of half the difference of its segments, when the perpendicular falls within; but as the cotangent of half the base to the co-tangent of half the sum of the segments, when the perpendicular falls without the triangle: And the sine of the sum of the two sides is to the sine of their difference as the co-tangent of half the angle contained by the sides, to the tangent of half the difference of the angles which the perpendicular makes with the same sides, when it falls within, or to the tangent of half the sum of these angles, when it falls without the triangle. If ABC be a spherical triangle, and AD a perpendicular to the base BC, sin (C+B): sin (C-B)::tan BC: tan (BD-DC), when AD falls within the triangle; but sin (C+B): sin (C-B):: cot BC: cot (BD+DC), when AC falls without. And again, |