small variation or fluxion of the equation on this hypothesis, it is AB. COS A. A sin c. BC + BC. cos c. c. Here, since we are ignorant of the magnitude of the errors or variations expressed by A and c, suppose them to be equal (a probable supposition, as they are both taken by the same instrument), and each denoted by : then will or, finally, BC = v sin A BC. AB sin c' COS C sin the equation will be c); BC (cot A cot c). This equation (in the use of which it must be recollected that v taken in seconds should be divided by £", that is, by the length of the radius expressed in seconds) gives the error BC in the estimation of BC occasioned by the errors in the angles A and c. Hence, that these errors, supposing them to be equal, may have ho influence on the determination of BC, we must have A = c, for in that case the second member of the equation will vanish. 2. But, as the two errors, denoted by A, and C, which we have supposed to be of the same kind, or in the same direction, may be committed in different directions, when the equation will be BC= v. BC (cot A + cot c); we must enquire what magnitude the angles A and C ought to have, so that the sum of their cotangents shall have the least value. possible; for in this state it is manifest that BC will have its least value. But, by the formulæ in chap. 3, we have cot Acot c = sin (A + c) sin A. sin c 2 sin B COS (AC) + cos B' Consequently, BC = ± V.EC. 2 sin B And hence, whatever be the magnitude of the angle B, the error in the value of BC will be the least when cos (A 2 C) is the greatest possible, which is, when a = c. We may therefore infer, for a general rule, that the most advantageous state of a triangle, when we would determine one side only, is when the base is equal to the side sought. 3. Since, by this rule, the base should be equal to the side sought, it is evident that when we would determine two sides, the most advantageous condition of a triangle is that it be equilateral. 4. It rarely happens, however, that a base can be commodiously measured which is as long as the sides sought. Supposing, therefore, that the length of the base is limited, but that its direction at least may be chosen at pleasure, we proceed to enquire what that direction should be, in the case where one only of the other two sides of the triangles is to be determined. Let it be imagined, as before, that AB is the base of the triangle ABC, and BC the side required. It is proposed to find the least value of cot A cot c, when we cannot have a = c. Now, in the case where the negative sign obtains, we have This equation again manifestly indicates the equality of AB and BC, in circumstances where it is possible: but if AB and BC are constant, it is evident, from the form of the denominator of the last fraction, that the fraction itself will be the least, cot c the least, when sin B is a maximum, that is, when в 90°. or cot A 5. When the positive sign obtains, we have cot A+ cot c = √(BC2 — AB2 sin2 a) AB SIN A BC2 =cot A+B sin2 A 1). cot A + Here, the least value of the expression under the radical sign, is obviously when A = 90°. And in that case the first term, cot A, would disappear. Therefore the least value of cot A + cot C, obtains when A = 90°; conformably to the rule giyen by M. Bouguer (Fig. de la Terre, pa. 88). But we have already seen that in the case of cot A cot C, we must have B = 90. Whence we conclude, since the conditions A =90°, B = 90°, cannot obtain simultaneously, that a medium result would give A = B. -- If we apply to the side Ac the same reasoning as to BC, similar results will be obtained: therefore in general, when the base cannot be equal to one or to both the sides required, the most advantageous condition of the triangle is, that the base be the longest possible, and that the two angles at the base be equal, These equal angles however, should never, if possible, be less than 23 degrees. PROBLEM 11. To deduce, from Angles measured Out of one of the stations, but Near it, the True Angles at the station. When the centre of the instrument cannot be placed in the vertical line occupied by the axis of a signal, the angles observed must undergo a reduction, according to circumstances. 1. Let 1. Let c be the centre of the station, P the place of the centre of the instrument, or the summit of the observed angle APB: it is required to find c, the measure of ACB, supposing there to be known APB = P, BPC = P, CP = d, BCL, ACR. Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles (th. 16 Geom.), we have, with respect to the triangle FAP, AIB = P+IAP; and with regard to the triangle BIC, AIB = c + CBP. Making these two values of AIB.equal, and transposing IAP, there results And, as the angles CAP, CBP, are, by the hypothesis of the problem, always very small, their sines may be substituted for their arcs or measures: therefore The use of this formula cannot in any case be embarrassing, provided the signs of sin p, and sin (P + p) be attended to. Thus, the first term of the correction will be positive, if the angle (P + p) is comprised between 0 and 180°; and it will become negative, if that angle surpass 180°. The contrary will obtain in the same circumstances with regard to the second term, which answers to the angle of direction p. The letter R denotes the distance of the object A to the right, L the distance of the object B situated to the left, and p the angle at the place of observation, between the centre of the station and the object to the left. 2. An approximate reduction to the centre may indeed be obtained by a single term; but it is not quite so correct as the form above. For, by reducing the two fractions in the second member of the last equation but one to a common denominator, the correction becomes 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 = Hence we manifestly have R sin A S. (A+P)* Which, by taking the expanded expressions for sin (P + p), and sin (AP), and reducing to seconds, gives C - P= d sin P. sin (A-p) 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 d. sin (p+p) R. sin 1" C-P= d. sin p P = The first of these will apply when the object A is a heavenly body, the second when B is one. When both A and B are such, then c P = 0. But without supposing either A or B infinite, we may have C - P = 0, or c = P 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 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 (theor. 54 Geom.) 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, Pm, be measured. Bisect mm' at the point n; and, placing there a signal T P staff, staff, measure the angle nPB, which, (since pn prolonged ob viously 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 rcd in the former investigation. If the circumference of the tower cannot be measured, and the radius thence inferred, proceed thus: Measure the angles BPT, 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. * B 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 Pmm' is equilateral, and PC = mm/3. T M PROBLEM III. 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, zó, 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 from this construction, that the angle z, of the spherical triangle zab, is equal to the horizontal angle a'c'B; and that, to find it, the three sides za = d, zb = d', abc, are given. Call the sum of theses; then the resulting formulæ of prob. 2 ch. iv, applied to the present instance, becomes sin 42 = sin c = √ sin † (s – d). sin {(s—d') sin d. sin d' |