3.5.7

2.4.6-8.9.29

+ &c. =523598775 =

of the

.12

circumference of a circle to the radius 1.

Again, the differential expression for the tangent is

Here, expanding the factor (1+tan? B), by the binomial theorem, and integrating each term of the result, there will arise

B = tan B tan3 B+tans B

tan7 B+ &c. Thus if в = 45°, then (ch. ii. pr. 6.) tan в = 1; consequently, substituting this value of tan B

Note. This series obviously converges very slowly: Euler (in his Analysis Infinitorum, vol. i.), and Dr. Hutton (in his 8vo. Tracts, vol. i.) have transformed it into others of rapid convergence..

SCHOLIUM. otolaryan, wit

Many curious, and indeed useful, results are deduci, ble from series of this kind, and of the opposite class, in which the sine, cosine, &c. is expressed in terms of the arc itself (ch. iv. equa. v.) As a specimen we select the following, first discovered by John Bernoulli.

The expression

+ &c.

sin x = x - +
2.3 2-3-4-5

when sin x = 0, becomes, after dividing by x,

Now, it is shown by writers on algebra, that the sum of the roots of every equation of this form is equal to the co-efficient of the 2d term with its sign changed;

that is in the present case =

Ꮖ

It is known also (chap.

iv. art. 3.) that the values of x which answer to the case of sin x = O, are TM, 2, 3, 4%, &c. * denoting the semicircumference to the radius 1. These successive values

of x or of being substituted, their results

In Geodesic operations, it is required to deduce from angles measured out of one of the stations, but near to it, the true angles at the station.

When the centre of the instrument with which horizontal angles are taken, cannot be placed in the vertical line occupied by the axis of the signal, the angles observed must undergo a reduction, according to circum

stances.

1. Let c be the centre of the sta- BA tion, 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, BC — L, ‚AC — R.

C P

A

Since the exterior angle of a triangle is equal to the sum of the two interior opposite angles, we have, with respect to the triangle IAP, 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 AP 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

When both A

The first of these will apply when the object A is a heavenly body, the second when в is one. and B are such, then c→ P = 0.

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 BẠC = 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-staff, measure the angle noв, which, (since Pn prolonged obviously passes through c the centre,) will be the angle p of the preceding investigation. Also, the distance rs added to the radius cs of the tower, will give rc = d 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 BPCĄ (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 PC = m'm √3.