And because P is always very nearly equal to c, the sine of A+ P will differ extremely little from sin (a + c), and may therefore be substituted for it, making L = Which, by taking the expanded expressions for (P + p); and sin (AP), and reducing to seconds, gives 3. When either of the distances R, L, becomes infinite, with respect to d, the corresponding term in the expression art. 1 of this problem, vanishes, and we have accordingly The first of these will apply when the object A is a heavenly Body, the second when в is one. When both A and B are such, then c But without supposing either A or B infinite, we may have c P0, or cp in innumerable instances.: that is, in all cases in which the centre P of the instrument is placed in the circumference of the circle that passes through the three points A, B, C; or when the angle BPC is equal to the angle BAC, or to BAC + 180°. Whence, though c should be inaccessible, the angle ACB may commonly be obtained by observation, without any computation. It may further be observed, that when P falls in the circumference of the circle passing L through the three points A, B, C, the angles A, B, C, may be determined solely by measuring the angles APB and BPC. For, the opposite angles ABC, APC, of the quadrangle inscribed in a circle, are = 180°. Consequently, ABC 180°- APC, and BAC = 180° (ABC + ACB) = 180° (ABC + APB). 4. If one of the objects, viewed from a further station, be a vane or staff in the centre of a steeple, it will frequently happen that such object, when the observer comes near it, is both invisible and inaccessible. Still there are various methods of finding the exact angle at c. Suppose, for example, the signal-staff be in the centre of a circular tower, and that the angle APB was taken at p near its base. Let the tangents PT, PT', be marked, and on them two equal and arbitrary distances Pm, rm, be measured. Bisect mm' at the point n; and, placing there a signal-staff, measure the angle nPB, which, (since rn prolonged obviously passes through c the centre,) will be the angle p of the preceding investigation. Also, the distance PS added to the radius cs of the tower, will give PC = d in the former investigation. If the circumference of the tower cannot be measured, and the radius thence inferred, proceed thus: Measure the angles BFT, BPT', then will BPC = (BPT +BPT)=p; and CPT: =BPT BPC: Measure PT, then PC = PT. Sec CPT=d. With the values of p and d, 'thus obtained, proceed as before. 5. If the base of the tower be polygonal and regular, as most commonly happens; assume P in the point of intersection of two of the sides prolonged, and BPC′ = } (BPT+BPT) as before, PT = the distance from P to the middle of one of the sides whose prolongation *passes through P; and hence PC is found, as above. If the figure be a regular hexagon, then the triangle rmm' is equilateral, and rc = mm 3. PROBLEM VII. To reduce angles measured in a plane inclined to the horizon, to the corresponding angles in the horizontal plane. Let BCA be an angle measured in a plane inclined to the horizon, and let B'CA' be the corresponding angle in the horizontal plane. Let d and d' be the zenith distances, or the complements of the angles of elevation ACA', BCB'. Then from z the zenith of the observer, or of the angle c, draw the arcs za, zb, of vertical circles, measuring the zenith distances d, d', and draw the arc ab of another great circle to measure the angle c. It follows C from this construction, that the angle z, of the spherical triangle zab, is equal to the horizontal angle A'CB; and -that, to find it, the three sides zad, zb d', ab = c, are given. Call the sum of these s; then the corresponding formula of ch. vi. pa. 90, applied to the present instance, becomes = B 'sin § (s — d) gin § (s — ď′) ̧. A 'A If h and h' represent the angles of altitude ACA', BCB', the preceding expression will become sin c√ sin § (c + h — h′) sin § (c + h' — h) ̧ cos h cos h' Or, in logarithms, log sin c(20+ log sin (c+hh') + log sin } (c+hh) log cos h-log cos h'). Cor. 1. If h=h', then is sin c = sin ACB cos h ; and og sin ACB10+ log sin ACB log cos h. Cor. 2. If the angles h and h' be very small, and nearly equal; then, since the cosines of small angles vary extremely slowly, we may, without sensible error, take = log sin A'CB 10+ log sin ACB log cos (h + h'). Cor. 3. In this case the correction x = A'CB' may be found by the expression ACB, ≈ = sin 1′′ (tan ¿c (10 − 4+ d′)2 — cot ¿c (d–d)o): d d' And in this formula, as well as the first given for sin c, d and d may be either one or both greater or less than a quadrant; that is, the equations will obtain whether ACA and BCB' be each an elevation or a depression. Scholium. By means of this problem, if the altitude of a hill be found barometrically, according to one of the methods described in Gregory's Mechanics, vol. i. book 5, or geometrically, according to some of those described in heights and distances, (chap. v.); then, finding the angles formed at the place of observation, by any objects in the country below, and their respective angles of depression, their horizontal angles, and thence their distances, may be found, and their relative places fixed in a map of the country; taking care to have a sufficient number of angles between intersecting lines, to verify the operations. PROBLEM VIII. In any spherical triangle, knowing two sides and the included angle; it is required to find the angle comprehended by the chords of those two sides. A Let the angles of the spherical triangle be A, B, C, the corresponding angles included by the chords A', B', C'; the spherical sides opposite the former, a, b, c; the chords respectively opposite the latter, a, 6, y; then, there are given b, c, and a, to find a'. a Here, from equa. 2, chap. vi. we have But cos c = cos (c + art. 23, chap. iv.) = (1 1-2 c) = cos2 c sin2 c (by sin c) sin dc sinc) sinc sin2 c. And in like manner cos a = 1 cos b = 1 2 sin2 36. tion becomes - 2 sin2 a, and Therefore the preceding equa cos b sin c cos c cos a + 12 sina: = 4 sin But sin a 2 sin2 c). Lα, sin b == 3, sin c =: which values substituted in the equation, we obtain, after a little re duction, Now, (equa. 11, chap. iv), cos A' = fore, by substitution, By cos A' By cos cos c cos A+ 322; whence, dividing by By, there results cos A cosb cos c cos a + 16 1v; or, lastly, by restoring the values of 3, 7, we have cos Acos cos c cos A+ sin 16 sin c....... (1.) Cor. 1. It follows evidently from this formula, that when the spherical angle is right or obtuse, it is always greater than the corresponding angle of the chords. Cor. 2. The spherical angle, if acute, is less than the' corresponding angle of the chords, when we have cos A sin sin c greater than 1 - costo cos c PROBLEM IX. Knowing two sides and the included angle of a rectilinear triangle, it is required to find the spherical angle of the two arcs of which those two sides are the chords. Here 6, 7, and the angle A' are given, to find A. Now, since in all cases, cos = √(! sin'), we have cos 6 cos c = √[(1 − sin2 ) (1 − sin2 c)]; we have also, as above, sin be, and si |