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 =
L. sin 1" or c

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'

If h and h' represent the angles of altitude ACA', BCB', the preceding expression will become

cos h. cos h'

Or, in logarithms,

log sin c (20+ log sin (c+hh') + log sin (c+h'- h) -log cos h-log cos h').

Cor. 1. If hh', then is sin c =

log sin A'CB' = 10 + log sin ACB —

sin ACE

cos h

-; and

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 ACB = 10 + log sin ACB log cos (h + h').

Cor. 3. In this case the correction

be found by the expression

x = sin 1′′ (tan ÷c(40 – d+d)

A'CB'-ACE, may

cot C

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 the method described in the 2d volume, or geometrically according to some of those described in heights and distances, or that given in the following problem; 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 IV.

Given the Angles of Elevation of any Distant object, taken at Three places in a Horizontal Right Line, which does not pass through the point directly below the object; and the Respective Distances between the stations; to find the Height of the Object, and its Distance from either station.

Let AED be the horizontal plane: FE the perpendicular height of the object F above that plane; A, B, C, the three places of observation; FAE, FBE, FCE, the respective angles

of

of elevation, and AB, BC, the given distances. Then, since the triangles AEF, BEF, CEF, are all right angled at E, the distances AE, BE, CE, will manifestly be as the cotangents of the angles of elevation at A, B, and c: and we have to determine the point E, so that those lines may have that ratio. To effect this geometrically use the following

Construction. Take BM, on AC produced, equal to BC, BN equal to AB; and make

MG BM(BC): cot A cot B,

and BN(AB) NG: cot B: cot c.

With the lines MN, MG, NG, constitute the triangle MNG; and join BG. Draw AE So, that the angle EAB may be equal to MGB; this line will meet BG produced in E, the point in the horizontal plane falling perpendicularly below F.

Demonstration. By the similar triangles AEB, GMB, we have AE BE:: MG MB:: cot A cot B, and BE BA(BN):: BM: BG.

Therefore the triangles BEC, BGN, are similar; consequently Whence it is obvious

BE EC: BN NG :: cot B: cot c.

that AE, BE, CE, are respectively as cot A, cot B, cot C.

Calculation. In the triangle MGN, all the sides are given, to find the angle GMN angle AEB. Then, in the triangle MGB, two sides and the included angle are given, to find the angle MGB angle EAB. Hence, in the triangle AEB, are known AB and all the angles, to find AE, and BE. And then

EF AE. tan A BE. tan B.

Otherwise, independent of the construction, thus.

Put ABD, BC = d, EF = x; and then express algebraically the following theorem, given at p. 128 Simpson's Select Exercises:

AE. BC CE'. AB BE2. AC + AC AB. BC, the line EB being drawn from the vertex E of the triangle ACE, to any point B in the base. The equation thence originating is

dr2. cot A+ Dr2. cot2c = (D+d).x2. cot2 B+(D+d)Dd. And from this, by transposing all the unknown terms to one side, and extracting the root, there results

(D+ d) Dd.

x = √ d. cot2 A+D . cot2 c− (D+ d) cot2 x'

Whence