TRIGONOMETRY

Trigono- THE word Trigonometry, from the Greek τρίγωνον and metry. μετρεσθαι, signifies the measurement of trigons or triangles. In general, a trigon is determined when three of its dimensions are known; the other dimensions may be computed from these data, and hence the complete measurement of a trigon includes the computation of those of its lines and angles which have not been ascertained by direct measurement. Now, this computation presents much greater difficulties than the operation of measuring does, and thus a treatise on trigonometry comes to be, virtually, a treatise on the art of computing the unknown parts of a trigon.

The ancient geometers sought to connect the numerical expression for an angle, with the rate of divergence of its sides, by cutting off from these two equal distances of a determinate length, and by ascertaining the value of the line joining their extremities. Since the extremities of these distances lie in the circumference of a circle, the connecting lines are the chords of the intercepted arcs; they were accordingly entered in tables as such.

The Arabian mathematicians, however, introduced tables of the half chords of the double arcs, to which they gave the name jeib, signifying in the bosom (whence the word jeb, a purse as carried in the bosom), which was translated into the Latin sinus, having the same signification. They also used tables of the lengths of the shadows of gnomons corresponding to various degrees of zenith distance, which are called by us tangents, but by them zyll, literally shadows. To these there have been added, in later times, tables of secants, or lines joining the vertex of the angle with the extremities of the tangents, in Arabic katy, that which cuts.

Almost all the operations of trigonometry are carried on by help of these tables of sines, tangents, secants, or, since the date of Napier's famous invention, by means of tables of their logarithms.

The early trigonometers used a radius consisting of sixty parts, but we now regard the radius of the circle as unit, and divide it decimally—that is, into ten, one hundred, one thousand equal parts, and so on; the values of the sines, tangents, secants, being expressed in decimal fractions of the radius. Thus, (fig. 1), AOB being an angle at the centre of the circle described with the radius OA, the arc AB is homologous to it, so that if the whole circumference be divided into any number of degrees, say 360, the arc AB contains as many of those divisions as the angle AOB contains of 360th parts of the entire revolution. Having let fall from B the line BC perpendicular to OA, BC is the sine of the arc AB, or of the angle AOB; also, having drawn AD to touch the circle at A, and to meet the continuation of OB in D, AD is the tangent, OD the secant, and AC the versed sine of the same arc: these also, by an extension of the meaning of the words, are said to be the tangent, secant, and versed sine of the angle AOB to the radius OA; and if OA be taken as unit, we write, indifferently,

The defect of an acute angle from a right angle, or of

its homologous are from a quadrant, is called the complement of that angle or of that arc, and, for shortness sake, the sine, tangent, and secant of the complement are called the cosine, cotangent, cosecant of the angle or arc. If, then, OE be drawn perpendicular to OA, the angle BOE is the complement of AOB, the arc BE the complement of AB; while FB, EG, OG, EF, which are the sine, tangent, secant, versed sine of BE, are called the cosine, cotangent, cosecant, covered sine of AB; and, conversely, CB, AD, OD, AC are the cosine, cotangent, cosecant, covered sine EB.

Among these lines there are several very obvious relations; thus, putting $\backslash alpha$ for the arc AB,

If one of the functions of an angle be given, we can, by help of these propositions, easily compute the values of the others. Thus, if $\backslash alpha$ be an angle such that its sine is

13th of the radius, we have $\backslash sin\; \backslash alpha^2\; =\; \backslash frac\{25\}\{169\}$, whence $\backslash cos^2\; \backslash alpha\; =\; \backslash frac\{144\}\{169\}$; $\backslash cos\; \backslash alpha\; =\; \backslash frac\{12\}\{13\}$; $\backslash tan\; \backslash alpha\; =\; \backslash frac\{\backslash sin\; \backslash alpha\}\{\backslash cos\; \backslash alpha\}\; =\; \backslash frac\{5\}\{12\}$; $\backslash cot\; \backslash alpha\; =\; \backslash frac\{1\}\{\backslash tan\; \backslash alpha\}\; =\; \backslash frac\{12\}\{5\}$; $\backslash sec\; \backslash alpha\; =\; \backslash frac\{1\}\{\backslash cos\; \backslash alpha\}\; =\; \backslash frac\{13\}\{12\}$ and $\backslash csc\; \backslash alpha\; =\; \backslash frac\{1\}\{\backslash sin\; \backslash alpha\}\; =\; \backslash frac\{13\}\{5\}$; $\backslash text\{ver\; \}\; \backslash alpha\; =\; \backslash frac\{1\}\{13\}$; $\backslash text\{cov\; \}\; \backslash alpha\; =\; \backslash frac\{8\}\{13\}$.

But the real problems which lie at the root of all trigonometrical researches are these: having given the sine, or any other of these functions of an angle, to compute of how many degrees that angle consists; and conversely, having given, numerically, the value of an angle to find the values of its various functions. The practical solution of these problems is to construct a trigonometrical canon in which the values of the sines, tangents, and secants are set down for every minute, second, or other small portion of the circumference.

If the sines and cosines of two arcs be known, we can readily obtain those of their sum and of their difference. Thus, let AB, BC (fig. 2), be two arcs, of which the sines BD, CE, and consequently the cosines OD, OE, are known: we may obtain the sine CF of their sum AC, or CF' of their difference AC' by the following process.

Through E draw EG parallel to BD and HEH' perpendicular to it; then, since the trigons ODB, OGE, CHE are similar, we have

$OB\; :\; OE\; ::\; BD\; :\; EG\; ::\; OD\; :\; OG$, or, putting $\backslash alpha$ for the arc AB, and $\backslash beta$ for BC or BC',
Again, $1\; :\; \backslash cos\; \backslash beta\; ::\; \backslash sin\; \backslash alpha\; :\; EG\; ::\; \backslash cos\; \backslash alpha\; :\; OG$; whence, $EG\; =\; \backslash sin\; \backslash alpha\; \backslash cdot\; \backslash cos\; \backslash beta$; $OG\; =\; \backslash cos\; \backslash alpha\; \backslash cdot\; \backslash cos\; \backslash beta$.
$OB\; :\; EC\; ::\; OD\; :\; CH\; ::\; DB\; :\; EH$; or, $1\; :\; \backslash sin\; \backslash beta\; ::\; \backslash cos\; \backslash alpha\; :\; CH\; ::\; \backslash sin\; \backslash alpha\; :\; EH$; whence, $CH\; =\; \backslash cos\; \backslash alpha\; \backslash cdot\; \backslash sin\; \backslash beta$; $EH\; =\; \backslash sin\; \backslash alpha\; \backslash cdot\; \backslash sin\; \backslash beta$.
Now, $FC\; =\; GE\; +\; CH$; $FC\text{'}\; =\; GE\; -\; CH$; $OF\; =\; OG\; -\; EH$; $OF\text{'}\; =\; OG\; +\; EH$; wherefore,

If we can by any means obtain the sine of one degree, equations (1) and (3) enable us, by putting $b=1^\backslash circ$, and $a$ successively equal to 1, 2, 3, &c., degrees, to obtain the sines and cosines of every degree in the half quadrant; and so for any other division: the essential matter, then, is to obtain the sine of the smallest sub-division which we intend to use.

On putting $b=a$, equations (1) and (3) become

and, observing that $\backslash sin^2\; a\; +\; \backslash cos^2\; a\; =\; 1$, the latter of these may be put under the forms

by help of which we can compute the sine and cosine of an arc when the cosine of its double is known.

Again, on putting $b=2a$, equations (1) and (3) become, after transformations,

by means of which the sine and cosine of the third part of an arc may be computed.

Similarly, on putting $3a$ for $a$, $2a$ for $b$, in the fundamental equations we obtain

which enable us to compute the sine and cosine of the fifth part of an arc.

In the ancient system of graduation, still very generally adhered to, the quadrant is divided into 90 degrees, the degree into 60 minuti primi or minutes; the minute again into 60 minuti secundi or seconds; while, in the modern or centesimal system the quadrant is divided into 100 degrees, the degree into 100 minutes, the minute into 100 seconds. The former graduation is accomplished by means of the prime divisors 2, 3, 5, while the modern requires only the use of 2 and 5, so that the above equations contain all that is needed for the formation of the canon of sines according to either of the systems.

The tri-section of an angle implies the solution of an equation of the third degree irreducible by Cardan's rule. The quinquisection requires the solution of an equation of the fifth degree. As no practicable method for reducing such equations was then known, the computers of the actual tables of sines were forced to use only repeated bisections; these bisections they carried on until they arrived at arcs so small as to be proportional to their sines without sensible error, and thence, by a common proportion, they found the sine of one minute. This process, besides being indirect, is attended with a serious practical inconvenience when the arc to be bisected is small. The cosine of a

small angle is nearly equal to unit, so that the quantity $\backslash frac\{1\}\{2\}\; -\; \backslash frac\{1\}\{2\}\; \backslash cos\; 2a$ comes to have few effective figures, and can only give the sine of the angle $a$ true to a small number of places. This circumstance compels us to carry the primary computations to a great number of decimal places. By employing the higher equations we avoid these inconveniences, as is seen from the following computation of the sine of one degree of the ancient division.

Since the chord of $60^\backslash circ$ is just equal to the radius, the sine of $30^\backslash circ$ must be $\backslash frac\{1\}\{2\}$, whence

Observing that, according to equations (9) (10)—

we obtain

By putting $x$ for $\backslash sin\; 5^\backslash circ$ in equation (11), we have

whence the following computation:—

which gives $\backslash sin\; 5^\backslash circ\; =\; .087\; 155\; 742$.

Similarly from equation (12) we have, putting $y$ for $\backslash cos\; 5^\backslash circ$,

whence

which gives $\backslash cos\; 5^\backslash circ\; =\; .996\; 194\; 698$.

The sine and cosine of one degree are now to be found by help of equations (13) (14). Putting $x$ for $\backslash sin\; 1^\backslash circ$, equation (13) becomes

whence the following very rapid computation:—

If we now, in order to compute the sine and cosine of one minute, proceed by bisection, we have

The first of these equations is clearly unfit for giving a result with great precision, for which reason, if our intention had been to compute the sine of one minute, it would have been better to have gone on with the bisection of $15^\backslash circ$; and then of $7^\backslash circ\; 30\text{'}$, so as to obtain the sine and cosine of $3^\backslash circ\; 45\text{'}$; thence by two trisections to reach $25\text{'}$; and, lastly, by two quinquisections to have obtained the sine and cosine of one minute.

The construction of the canon of sines is greatly facilitated by employing the method of differences. Thus, if we take three equi-different arcs, $a\; -\; b$, $a$, $a\; +\; b$, their sines with the first and second differences stand thus,

and this second difference may, by help of equations (1) and (2), be put under the form $-2\backslash sin\; a(1\; -\; \backslash cos\; b)\; =\; -2\backslash sin\; a\; \backslash cdot\; \backslash text\{ver\; \}\; b$. Now, since $b$ is necessarily a small angle, its versed sine, that is, the defect of its cosine from radius, must also be very small, so that the multiplication by it is attended with little labour. Again, if $a\; -\; 2b$, $a\; -\; b$, $a$, $a\; +\; b$, $a\; +\; 2b$, be five equidifferent arcs, the fourth difference of their sines is

which may be put under the forms

And, further, if we take seven equidifferent arcs from $a\; -\; 3b$ to $a\; +\; 3b$, the sixth difference of their sines is

and this law of formation extends to all differences of an even order.

But in applying the method of successive differences, the minute errors unavoidable in the last placed figures would go on accumulating, wherefore we must provide periodical tests of the accuracy of the work. Now, we have already obtained the sines of $15^\backslash circ$, $30^\backslash circ$, $60^\backslash circ$, $75^\backslash circ$, wherefore the sine of $45^\backslash circ$ only is wanted to complete the series of arcs differing by $15^\backslash circ$. But of $45^\backslash circ$ the sine is, necessarily, equal to the cosine, and therefore

so that our first table of sines is this:

The equation $\backslash sin\; 2a\; =\; 2\; \backslash sin\; a\; \backslash cdot\; \backslash cos\; a$ may be put in the form $\backslash sin\; 2a\; =\; 2\; \backslash sin\; a\; \backslash cdot\; \backslash sin\; a\; \backslash cdot\; 2\; \backslash text\{\; ver\; \}\; a$, which, especially when $a$ is a small arc, is more convenient for computation. Putting $a\; =\; 5^\backslash circ$, this equation becomes

whence the following table is readily constructed:

Having written the sines of $0^\backslash circ$ and of $5^\backslash circ$, we get the first differences; and the first of the second differences is obtained by taking the product of $\backslash sin\; 5^\backslash circ$ by $-2\; \backslash text\{\; ver\; \}\; 5^\backslash circ$, from which the second of the first differences and the sine of $10^\backslash circ$ are had. The first of the third differences, viz., 658 260 is the product of the second of the first differences, viz., + 86 492 435 by the same multiplier $-2\; \backslash text\{\; ver\; \}\; 5^\backslash circ$; from it we get the next second difference, the next first difference, and the sine of $15^\backslash circ$, the coincidence of which with the previously determined value shows the work to have been accurately done. The first of the fourth differences, 10 059, is the product of $-1321\; 568$ by the same multiplier, or, as is preferable, of $\backslash sin\; 10^\backslash circ$ by $(2\; \backslash text\{\; ver\; \}\; 5^\backslash circ)^2$. Each successive fourth difference is the product of the same multiplier $(2\; \backslash text\{\; ver\; \}\; 5^\backslash circ)^2$ by the sine of the next arc; now this multiplier .000 057 921 has few effective figures, and thus the labour of the calculation is much reduced.

The same process may be extended to the sixth order of differences, which are obtained by multiplying the columns of sines by $(-2\; \backslash text\{\; ver\; \}\; 5^\backslash circ)^3\; =\; .000\backslash \; 000\backslash \; 441$, which has still fewer effective figures; but the advantage obtained by pushing the differences to a high order is counteracted by

Trigono-
metry. the circumstance, that the minute errors, unavoidable in the last place figures, accumulate to cause inconvenience.

When the subdivision is more minute, this method is still more rapid; thus, in constructing the table of the sines of arcs differing by one degree, we use the multipliers

It is always necessary to carry the computation of the differences to two or three places beyond what is intended to be used. This is well seen in the above example, in which the values of the sines are given true to the nearest figure in the ninth decimal place: the fourth differences found by help of these correct values, are 8, 1; whereas, by computation from the sines of $2^\backslash circ$ and $3^\backslash circ$, they ought to have been 3.25 and 4.86, if the work had been carried two places farther.

This illustration is sufficient to exemplify the manner in which trigonometrical tables are constructed. Various contrivances are employed for diminishing the labour of making tables of secants, tangents, and logarithmic sines. A complete account of these would extend this article to an unreasonable length; we shall, therefore, at once address ourselves to the method of using those tables which have been already constructed.

Hitherto we have only considered the sines, tangents, and secants of arcs within the quadrant; and must now examine the cases of arcs extending to the whole circumference, or even beyond it.

Let a point start from A, and move round the circumference of the circle described from the centre O with the radius OA (fig. 3); when the moving point has passed over the arc AB, the sine has grown from zero to CB; as the arc continues to increase, the sine also increases, until the arc becomes a complete quadrant AD, at which time the sine is the radius OD. After the arc has passed into the second quadrant, the sine FE shortens to become zero, when the arc is the semicircumference AG; and

it is obvious that the sine of the arc AE, which is greater than the quadrant, is also the sine of its supplement GE, which is as much less than the quadrant.

When the arc, as AH, exceeds the semicircumference, its sine IH appears on the other side of the diameter AOG, and this position is represented by the sign $-$. The sine of AK (three quadrants) is the radius OK, and is therefore represented by $-1$; and when the arc, as AL, exceeds three quadrants, its sine ML, actually decreasing, is, in algebraic language, said to be increasing, that is, becoming less subtractive. The sine of the whole circumference AA is again zero; and if the moving point be supposed to continue its progress, the same changes in the value of the sine are repeated in each succeeding revolution. The cosine of the arc undergoes corresponding changes; it is positive in the first quadrant, negative in the second and third, and again positive in the fourth quadrant.

Resolution of Right-angled Triangles.

In a right-angled triangle ABC, besides the right angle at C, we have one angle and three sides to consider: of these, any two being given, the other two may be computed thus:—

CASE 1.—An angle and the hypotenuse being given.

If the hypotenuse AB were made the radius, BC would be the

sine, and AC the cosine of the angle A; hence $BC\; =\; AB\; \backslash cdot\; \backslash sin\; A$, $AC\; =\; AB\; \backslash cdot\; \backslash cos\; A$ (fig. 4).

Let A be $27^\backslash circ\; 53\text{'}$; $AB\; =\; 572.8$; we obtain BC and AC thus:

CASE 2.—An angle and the adjacent side being given.

If the given side, say AC, were made the radius, CB would be the tangent, and AB the secant of the adjacent angle A; hence $CB\; =\; AC\; \backslash cdot\; \backslash tan\; A$, $AB\; =\; AC\; \backslash cdot\; \backslash sec\; A$.

Example.—Let A = $57^\backslash circ\; 41\text{'}$, $AC\; =\; 897.7$

CASE 3.—An angle and the side opposite to it being given.

If BC were made the radius, CA would be the cotangent, AB the cosecant of the opposite angle A; whence $CA\; =\; BC\; \backslash cdot\; \backslash cot\; A$, $AB\; =\; BC\; \backslash cdot\; \backslash csc\; A$.

Thus, let $BC\; =\; 419.72$; $A\; =\; 15^\backslash circ\; 31\text{'}$, and the calculation may be arranged as under:

In these three examples the logarithm of the given side is placed in the middle, so as to be conveniently added to the logarithm above for the one result, to the logarithm below for the other.

CASE 4.—The hypotenuse and one side being given.

Example.—Let $AB\; =\; 894.37$; $BC\; =\; 514.63$; then when the angle as well as the other side is wanted, the computation may be arranged thus:—

Here, from the logarithm of BC we subtract that of AC, in order to obtain the logarithmic sine of the angle A, and thence the angle A itself. Then, taking the logarithmic cosine of that angle, we write it above or adjacent to the logarithm of AB, the space having been purposely left open for it, these added together give the logarithm of AC.

When the side AC alone is wanted, this process is somewhat long; it is preferable to proceed as under:—

This operation is founded on the fact, that the difference between the squares of two lines is equivalent to the rectangle under the sum and the difference of those lines.

CASE 5.—When the two sides are given.

If AC = 980.91, CB = 762.43, the angle A and the hypotenuse AB may be found thus:—

Resolution of Oblique-Angled Triangles.

In general we have to consider two angles and the three sides of a triangle: of these, any three being given, the remaining two may be computed.

CASE 6.—Two angles and a side being given.

The third angle is found by subtracting the sum of the two given ones from 180°.

Having described a circle round the triangle ABC (fig. 5), join the centre O with each of the corners, and let fall the perpendiculars OD, OE, OF; then the angles BOD, COE, AOF are, respectively, equal to BAC, CBA, ACB. Now, BD is the sine of the angle BOD to the radius OB; so is CE of COE, and AF of AOF, so that the three lines BD, CE, AF, or their doubles BC, CA, AB, are proportional to the sines of the opposite angles BAC, CBA, ACB.

When both of the other sides are wanted, the computation can be arranged neatly, so as to avoid unnecessary figuring, by observing that the cosecant of an angle is the inverse of its sine.

Example.—Let AB = 1378.7; BAC = 47° 53'; ABC = 65° 19', and consequently ACB = 66° 44'.

The logarithms of the results are here obtained by adding together the three upper and the three under logarithms.

CASE 7.—Two sides and an angle opposite to one of them being given.

As, in certain circumstances, this case admits of a double solution, it can only be safely used when there are the means of discriminating between the two results. Thus if, in the measurement of some triangle, we had found CAB = 37° 27', AB = 3281, BC = 2960, it would have been difficult to determine which of the two results should be taken. For in finding the angle C by means of the proportion BC : AB :: sin A : sin C, we obtain for the log sine of C the value 9.999 9844, which is the log sine of 89° 30', and somewhere between 48° and 55°, and is also the log sine of the supplement, viz., 90° 29', and between 5° and 12°; the angle C may thus be either 89° 30' 50", or 90° 29' 10", and each of these values with an uncertainty of several seconds.

If we take the former value, we obtain 53° 02' 10" for the remaining angle B, and thence get the length of the third side AC, as follows:—

But if we take the latter value, we have C = 52° 03' 50", giving the following calculation for AC:—

Thus, it seems that this mode of measuring a triangle is to be avoided whenever the angle opposite to the other given side is nearly a right angle; both on account of the difficulty of discriminating between the two results, and of the inexactitude with which each of these is obtained.

When the side opposite the measured angle is the greater of the two, there can be no ambiguity.

CASE 8.—Two sides and the angle contained by them being given.

When the sides BA, BC, and their contained angle ABC, are known, the other angles and the third side may be computed by drawing a perpendicular from B to AC, and by resolving the right-angled triangles thus formed. The following process is, however, generally preferred:—

Having produced one side CB, lay off BD, BE, each equal to BA, join DA, EA, and through C draw CF parallel to EA. Then ABD, the supplement of ABC, is equal to the sum of the two angles BAC, BCA, wherefore BEA is half the sum of A and C, while EAC or ACF is half their difference. Also, EAD is a right angle, so that FD is the tangent of DCF, FA the tangent of FCA to the radius CF; but FD : FA :: CD : CH; wherefore,

By help of this proportion we can find the half difference of the two angles, and thence the angles A and B themselves. Afterwards the third side is usually computed by help of the law stated in Case 6; but the following process, which is believed to be new, enables us to avoid the seeking out of logarithms, and the opening of the trigonometrical canon at so many places.

In the triangle AEC we have $\backslash sin\; CAE\; :\; \backslash sin\; AEC\; ::\; EC\; :\; CA$; that is,

Or, in the triangle ADC, $\backslash sin\; DAC\; :\; \backslash sin\; ADC\; ::\; DC\; :\; CA$; that is,

By means of either of these proportions we can find the third side CA.

Example.—Let AB = 738.6; BC = 1079.3; ABC = 67° 42'; whence A + C = 112° 18'.

Trigono- When, as often happens, the third side only is wanted, the fol- metry. lowing process is perhaps the most expeditious:—

Through B draw BG parallel to CF, and, of course, bisecting DA. Then $DC^2\; -\; CA^2\; =\; DF^2\; -\; FA^2\; =\; 4\; DG\; \backslash cdot\; GF$, but $DG\; =\; AB\; \backslash cdot\; \backslash cos\; \backslash frac\{B\}\{2\}$.

$GF\; =\; BC\; \backslash cdot\; \backslash cos\; \backslash frac\{B\}\{2\}$; wherefore $AC^2\; =\; DC^2\; -\; 4\; AB\; \backslash cdot\; BC\; \backslash left(\; \backslash cos\; \backslash frac\{B\}\{2\}\; \backslash right)^2$. Hence, if we find M a mean proportional between DG and GF, we must have

$(AB\; +\; BC\; +\; 2M)$; whence the following computation:—

CASE 9.—The three sides being given.

Bisect the angles BAC, BCA, and BCR the supplement of BCA, by the lines AOQ, CO, and CQ (Fig. 7); draw also the perpendiculars OP, QR; then it is easy to show that O is the centre and OP the radius of the inscribed circle, that Q is the centre and QR the radius of one of the circles of external contact; also, that AR is the semiperimeter of the triangle, and AP, PC, CR, the excesses of that semiperimeter above the sides CB, BA, AC respectively.

Fig. 7.

The triangle QCR is similar to COP, wherefore $OP\; :\; PC\; ::\; CR\; :\; RQ$, and $OP\; \backslash cdot\; RQ\; =\; PC\; \backslash cdot\; CR$. Again, $AR\; :\; AP\; ::\; RQ\; :\; OP\; ::\; OP\; \backslash cdot\; RQ\; :\; OP^2$, wherefore $AR\; :\; AP\; ::\; PC\; \backslash cdot\; CR\; :\; OP^2$, or $OP\; =\; \backslash sqrt\{\backslash frac\{AP\; \backslash cdot\; PC\; \backslash cdot\; CR\}\{AR\}\}$. But $AP\; :\; PO\; ::\; R\; :\; \backslash tan\; \backslash frac\{1\}\{2\}\; A$, wherefore

$\backslash tan\; \backslash frac\{1\}\{2\}\; A\; =\; \backslash frac\{PO\}\{AP\}$; and similarly, $\backslash tan\; \backslash frac\{1\}\{2\}\; C\; =\; \backslash frac\{PO\}\{PC\}$, $\backslash tan\; \backslash frac\{1\}\{2\}\; B\; =\; \backslash frac\{PO\}\{CR\}$. Also, the area of the triangle is equivalent to the rectangle under AR and OP, or $ABC\; =\; \backslash sqrt\{AP\; \backslash cdot\; PC\; \backslash cdot\; CR\; \backslash cdot\; RA\}$. Hence, the three angles and the area can be found by a very concise operation.

Example.—Let $CB\; =\; 517.7$; $BA\; =\; 789.5$; $AC\; =\; 904.6$.

the last four logarithms being obtained by subtracting each of the first four from the logarithm of the inscribing radius.

SPHERICAL TRIGONOMETRY.

A spherical triangle is a portion of the surface of a sphere enclosed by three arcs of great circles, and represents the corner or solid angle formed by the meeting of three planes at the centre of the sphere. The sides of the triangle measure the angles, and the angles of the triangle measure the edges of the corner. Since the sum of the three angles of a spherical triangle exceeds half a revolution by a quantity proportional to the surface of the triangle, we have to

consider the three angles and the three sides, any three of which being given, the other three may have to be calculated; and therefore the cases are more numerous than in plane trigonometry. But the number of investigations is

Fig. 8.

reduced one-half by the consideration of what is called the supplemental or polar triangle.

From each of the corners A, B, C, of the spherical triangle ABC, as a pole describe a great circle, so as to form a new triangle PQR, then the corners of PQR are necessarily the poles of the sides of ABC.

Let the sides BA, BC, produced if necessary, meet PR in S and T, then the arc ST is homologous with the angle ABC; now PT and SR are quadrants, wherefore PR is the supplement of ST; wherefore each side of PQR is supplementary to an angle of ABC, and conversely. Hence, if one spherical triangle have its sides supplementary to the angles of another, its angles also are supplementary to the sides of that other.

Right-angled Spherical Triangles.

Let ABC represent a spherical triangle having a right angle at C; join A, B, C with O, the centre of the sphere,

Fig. 9.

so as to trace a corner bounded by the three planes AOB, BOC, COA, the two last being perpendicular to each other.

In OC take any point D, thence draw DF perpendicular to OA, at F in the plane AOB raise FE also perpendicular to OA, and join DE; DE is evidently normal to the plane AOC, and the angle DFE measures the inclination of the two planes COA, BOA, and is thus equal to the angle A.

If we suppose OE to be the tabular radius, OD is the cosine of BC, DE its sine; OF the cosine of BA, EF its sine; but $OD\; :\; OF\; ::\; R\; :\; \backslash cos\; AC$, wherefore $R\; :\; \backslash cos\; AC\; ::\; \backslash cos\; BC\; :\; \backslash cos\; AB\; (1)$; and again, $FE\; :\; ED\; ::\; R\; :\; \backslash sin\; CAB$, wherefore $R\; :\; \backslash sin\; A\; ::\; \backslash sin\; AB\; :\; \backslash sin\; BC\; (2)$.

If we suppose OD to be the tabular radius, DE becomes the tangent of BC, DF the sine of CA; but $FD\; :\; DE\; ::\; R\; :\; \backslash tan\; DFE$, wherefore $R\; :\; \backslash tan\; A\; ::\; \backslash sin\; AC\; :\; \backslash tan\; BC\; (3)$

Lastly, if OF be made the tabular radius, FE becomes the tangent of AB, FD that of AC; now $EF\; :\; FD\; ::\; R\; :\; \backslash cos\; EFD$, whence $R\; :\; \backslash cos\; A\; ::\; \backslash tan\; AB\; :\; \backslash tan\; AC$ (4).

As an analogous construction may be made for the angle B, we have similarly $R\; :\; \backslash sin\; B\; ::\; \backslash sin\; AB\; :\; \backslash sin\; AC$ (5); $R\; :\; \backslash tan\; B\; ::\; \backslash sin\; BC\; :\; \backslash tan\; AC$ (6); and $R\; :\; \backslash cos\; B\; ::\; \backslash tan\; AB\; :\; BC$ (7).

By inverting the 7th proportion, and combining it with the second, we obtain $\backslash cos\; B\; :\; \backslash sin\; A\; ::\; \backslash cos\; AB\; :\; \backslash cos\; BC$, wherefore $1\; :\; \backslash cos\; AC\; ::\; \backslash sin\; A\; :\; \backslash cos\; B$ (8); and similarly, $1\; :\; \backslash cos\; BC\; ::\; \backslash sin\; B\; :\; \backslash cos\; A$ (9); also, by combining the 1st, 3d, and 6th proportions, we have $1\; :\; \backslash cot\; A\; ::\; \backslash cot\; B\; :\; \backslash cos\; AB$ (10). From these ten proportions we have the following solutions of the various cases:—

CASE 1.—Given A and B.

CASE 2.—Given A and AB.

CASE 3.—Given A and AC.

CASE 4.—Given A and BC.

CASE 5.—Given AB and BC.

CASE 6.—Given AC and BC.

Quadrantal Spherical Triangles.

When one side of a spherical triangle is a quadrant, it may be revolved by considering its conjugate. Thus, if ABC be a triangle, having $AB\; =\; 90^\backslash circ$, we may form another triangle $A\text{'}B\text{'}C\text{'}$, the angles of which are the supplements of the sides of ABC, and therefore having $C\text{'}\; =\; 90^\backslash circ$. In this way, we obtain the ten following equations which contain the solution of every possible case.

Spherical Triangles in General.

CASE 1. Given the three sides.

Having made a construction analogous to that given in Case 9 of Plane Trigonometry, O is the pole of a small circle which touches

Fig. 10.

the three sides internally, and Q that of another small circle which touches the same three sides externally.

In the right-angled triangle OCP, we have $R\; :\; \backslash tan\; OCP\; ::\; \backslash sin\; CP\; :\; \backslash tan\; PO$; and similarly in QCR, $R\; :\; \backslash tan\; QCR\; ::\; \backslash sin\; CR\; :\; \backslash tan\; RQ$; but QCR is the complement of OCP, wherefore $\backslash tan\; OCP\; :\; R\; ::\; \backslash tan\; QCR$, wherefore $\backslash tan\; PO\; :\; \backslash sin\; CP\; ::\; \backslash sin\; CR\; :\; \backslash sin\; RQ$, or $\backslash tan\; PO\; \backslash cdot\; \backslash tan\; RQ\; =\; \backslash sin\; PC\; \backslash cdot\; \backslash sin\; CR$. Again, in the triangles APO, ARQ, $R\; :\; \backslash tan\; QAR\; ::\; \backslash sin\; AR\; :\; \backslash tan\; RQ\; ::\; \backslash sin\; AP\; :\; \backslash tan\; PO$, wherefore $\backslash sin\; AR\; :\; \backslash sin\; AP\; ::\; \backslash tan\; RQ\; :\; \backslash tan\; PO\; ::\; \backslash tan\; PO\; \backslash cdot\; \backslash tan\; RQ\; :\; \backslash tan\; PO^2$, or $\backslash sin\; AR\; :\; \backslash sin\; AP\; ::\; \backslash sin\; PC\; \backslash cdot\; \backslash sin\; CR\; :\; \backslash tan\; PO^2$, whence

The tangent of PO being thus found, we readily obtain the tangent

of the half angles by observing that $\backslash tan\; \backslash frac\{A\}\{2\}\; =\; \backslash frac\{\backslash tan\; PO\}\{\backslash sin\; AP\}$, $\backslash tan\; \backslash frac\{C\}\{2\}\; =$

$\backslash frac\{\backslash tan\; PO\}\{\backslash sin\; PC\}$, and $\backslash tan\; \backslash frac\{B\}\{2\}\; =\; \backslash frac\{\backslash tan\; PO\}\{\backslash sin\; CR\}$. Hence a form of procedure quite

similar to that for the analogous case of plane triangles.

Example.—Let $BC\; =\; 84^\backslash circ\; 27\text{'}\; 48\text{'}\text{'}$; $AB\; =\; 95^\backslash circ\; 44\text{'}\; 51\text{'}\text{'}$; $CA\; =\; 53^\backslash circ\; 14\text{'}\; 17\text{'}\text{'}$.

If only one of the angles had been wanted, say that at A, we should have used the formula

CASE 2. The three angles being given.

A, B, C, being the three given angles; let us construct a triangle $A\text{'}B\text{'}C\text{'}$, having $B\text{'}C\text{'}\; =\; 180^\backslash circ\; -\; A$; $C\text{'}A\text{'}\; =\; 180^\backslash circ\; -\; B$; $A\text{'}B\text{'}\; =\; 180^\backslash circ\; -\; C$; then shall we have $A\text{'}R\text{'}\; =\; 270^\backslash circ\; -\; \backslash frac\{1\}\{2\}(A+B+C)$; $A\text{'}P\text{'}\; =\; 90^\backslash circ\; -\; \backslash frac\{1\}\{2\}(-A+B+C)$; $P\text{'}C\text{'}\; =\; 90^\backslash circ\; -\; \backslash frac\{1\}\{2\}(A+B-C)$; $C\text{'}R\text{'}\; =\; 90^\backslash circ\; -\; \backslash frac\{1\}\{2\}(A-B$

$+\; C)$; wherefore $\backslash tan\; \backslash frac\{A\text{'}\}\{2\}\; =\; \backslash sqrt\{\backslash frac\{\backslash cos(S-C)\; \backslash cdot\; \backslash cos(S-B)\}\{-\backslash cos(S-A)\; \backslash cdot\; \backslash cos\; S\}\}$ in

which S is put for $\backslash frac\{1\}\{2\}(A+B+C)$. But $A\text{'}\; =\; 180^\backslash circ\; -\; BC$, where-

fore $\backslash cot\; \backslash frac\{BC\}\{2\}\; =\; \backslash sqrt\{\backslash frac\{\backslash cos(S-C)\; \backslash cdot\; \backslash cos(S-B)\}\{-\backslash cos(S-A)\; \backslash cdot\; \backslash cos\; S\}\}$, by help of which

formulas any one of the sides may readily be found. When all the three sides are wanted, an arrangement similar to the last may be used.

Example.—$A\; =\; 89^\backslash circ\; 58\text{'}\; 43\text{'}\text{'}$; $B\; =\; 76^\backslash circ\; 47\text{'}\; 19\text{'}\text{'}$; $C\; =\; 69^\backslash circ\; 19\text{'}\; 48\text{'}\text{'}$.

CASE 3. Two sides and the contained angle being given.

Let AB, BC, be the given sides, ABC the given angle. From A draw the arc AD of a great circle perpendicular to the side BC, then we have $\backslash tan\; BA\; \backslash cdot\; \backslash cos\; B\; =\; \backslash tan\; BD$, whence BD, and then DC can be found. Now, $R\; :\; \backslash tan\; B\; ::\; \backslash sin\; BD\; :\; \backslash tan\; DA$ and $\backslash tan\; C\; :\; R\; ::\; \backslash tan\; DA\; :\; \backslash sin\; DC$, wherefore, compounding $\backslash sin\; DC\; :\; \backslash sin\; BD\; ::\; \backslash tan\; B\; :\; \backslash tan\; C$, which gives us C; and, again, $R\; :\; \backslash cos\; AD\; ::\; \backslash cos$

Trigono- BD : cos AB; R : cos AD :: cos CD : cos AC, whence cos BD :
metry. cos DC :: cos AB : cos AC.

also tan KG = sin CK . tan KCG, wherefore sin BK : sin KC ::
tan KCG : tan KBG, that is

Also, since GDC is a right angle, cos CG = cot DGC . cot GCD,
while cos BG = cot BGK . cot GBK; wherefore since DGC = BGK,
cos CG : cos BG :: cot GCD : cot GBK :: tan GBK : tan GCD;
but cos CG : cos BG :: cos CK : cos BK, wherefore cos CK : cos
BK :: tan GBK : tan GCD, that is

Again, in the right-angled triangle GCK, tan GC . cos GCK =
tan KC; and in GDC, tan GC . cos GCD = tan CD; wherefore cos
GCK : cos GCD :: tan KC : tan CD, that is

Lastly sin GK = sin GC . sin GCK; sin GD = sin GC . sin GCD;
sin GK . tan BGK = tan BK and sin GD . tan DGC = tan DC;
wherefore, since BGK = DGC; sin GCK : sin GCD :: tan BK :
tan DC, that is

These four proportions are known under the name of Nepair's
Analogies, in honour of their illustrious discoverer. By help of
the first two we obtain half the difference and half the sum of the
unknown angles, whence those angles themselves can be found;
and then by help of either the third or the fourth we compute the
half of the third side.

Taking the preceding example, we have

Or, thus

AC in this example is the angular distance of the star $\backslash alpha$ Andro-
medæ from $\backslash beta$ Orionis (1850).

CASE 4.—Two angles and the intermediate side being
given.

The solution of this case is quite analogous to that of the pre-
ceding; thus, if the two angles at A and C and the interjacent side
AC be known, we may let fall from A a perpendicular AD upon
the opposite side, so as to form two right-angled triangles.

Then cos AC . tan C = cot CAD, whence CAD and, by subtrac-
tion, DAB can be found. Now in the right-angled triangle ADB,
tan AD = tan AB . cos DAB, while in ADC, tan AD = tan AC .
cos CAD, therefore cos DAB : cos CAD :: tan AB : tan AC.
From which AB is obtained. Again, cos B = cos AD . sin DAB and cos
C = cos AD . sin DAC, wherefore sin CAD : sin DAB :: cos C :
cos B which gives the angle B. Lastly, to find BC we have tan
AC . cos C = tan DC, tan DC . tan DAB . cot CAD = tan DB.

Fig. 31.

In order to find the angle A, we observe that cot ABD . sec BA
= tan BAD; and also, that, since tan BD = sin AD . tan BAD
and tan DC = sin AD, tan DAC, tan BD : tan DC :: tan BAD :
tan DAC.

Example.—Let AB = 61° 44' 14"; BC = 98° 22' 45";
ABC = 76° 40' 20".

This process is convenient when only the third side, or
only one of the angles, is wanted; for the complete solution,
the following investigation is to be preferred.

AB and BC being the given
sides, bisect the third side AC
in D, and make DF perpendicu-
lar to AC, continuing it to meet
AB produced in F, join CF,
thus forming an isosceles spher-
ical triangle AFC. Bisect the
angle BCF by CG, and let fall
the three perpendiculars GK,
GL, GM; these three are evi-
dently equal to each other, and
also BK to BM, wherefore the
arc BG must bisect the angle
FBC. Also, since MF = FL,
AM = CL = CK, so that CK is
half the sum, KB half the differ-
ence of AB and BC, while ACG
is half the sum, GCK half the
difference of the angles A and
C; moreover, since the whole
revolution at G is made up of 2 FGL, 2 LGC, and 2 BGK; the half
revolution is composed of FGL, LGC, BGK, therefore CGD is
equal to BGK; at the same time, it may be observed, that KBG is
the complement of the half of the angle ABC.

Fig. 12.

Since BKG is a right angle tan KG = sin BK . tan KBG; and
VOL. XXI.

Example.—Let $A\; =\; 107^\backslash circ\; 23\text{'}\; 46\text{'}\text{'}$; $C\; =\; 75^\backslash circ\; 49\text{'}\; 28\text{'}\text{'}$; $AC\; =\; 67^\backslash circ\; 29\text{'}\; 32\text{'}\text{'}$.

Or the solution may be accomplished by help of Napier's analogies thus;

Or thus for B:

CASE 5.—Two sides and an angle opposite one of them being given.

This, like the corresponding case in plane trigonometry, admits of two solutions, and we can only discriminate between these by a knowledge of the circumstances with which the work is connected.

We have, in relation to the figure for case 3, $R\; :\; \backslash sin\; B\; ::\; \backslash sin\; BA\; :\; \backslash sin\; AD$; $\backslash sin\; C\; ::\; \backslash sin\; CA\; :\; \backslash sin\; AD$; therefore $\backslash sin\; B\; :\; \backslash sin\; C\; ::\; \backslash sin\; AC\; :\; \backslash sin\; AB$, and in general $\backslash sin\; A\; :\; \backslash sin\; BC\; ::\; \backslash sin\; B\; :\; \backslash sin\; CA\; ::\; \backslash sin\; C\; :\; \backslash sin\; AB$.

By help of this proportion we can compute the sine of the other opposite angle, and thence obtain the two supplementary angles of which it is the sine. Having ascertained which of the two angles is to be taken, we have now two sides and the two angles opposite to them, and can readily obtain the third side and the third angle by help of Napier's analogies.

CASE 6.—Two angles and a side opposite one of them being given.

The very same remarks apply to this as to the preceding case, the modes of solution being so closely related that it is not worth while to give the details of both.

Example.—$A\; =\; 78^\backslash circ\; 21\text{'}\; 40\text{'}\text{'}$; $C\; =\; 59^\backslash circ\; 47\text{'}\; 30\text{'}\text{'}$; $AB\; =\; 48^\backslash circ\; 13\text{'}\; 20\text{'}\text{'}$; $BC$ obtuse.

Or thus for AC, and B:

In this account of trigonometrical calculations, the leading cases and the most convenient modes of operating have alone been given. For fuller information on special cases, and the formulae applicable to them, the reader is referred to any of the thousand and one treatises on the subject.

FIELD OPERATIONS.

The operations of the surveyor may be treated under three heads: the measurement of lines; the measurement of angles; and the determination of direction.

Graduated rods, tapes, and chains are used for the measurement of distance, the mode of operating depending on the degree of precision which is desired; for ordinary field-work, chains of 100 links are used, some being 100, some 65, and, for lightness, some only 50 feet long. The chain of 66 feet was contrived by Gunter in order to obtain the advantage of decimal calculation, for ten squares of such a chain make one acre, and thus the acre consists of 100,000 square links; 80 of Gunter's chains make one mile.

When the tape is used, care must be taken to stretch it always to the same tension; it is not to be depended on for nice work; its lightness is its great recommendation.

For the measurement of base-lines, where the utmost attainable degree of precision is needed, rods of wood or metal are employed, and means are provided for ascertaining their temperature in order that their expansions and contractions may be allowed for; the readings also are made by help of elaborate micrometric apparatus.

The measurement of angles proper is accomplished by means of the sextant, reflecting circle, repeating circle, and analogous instruments.

The repeating circle consists, essentially, of two telescopes, AB, CD, turning on either side of a graduated circle, and having the planes in which they move brought as closely as possible together. Each of these telescopes can be secured and adjusted to any part of the limb by means of a clamp and tangent screw; and one of them, at least, as AB, carries a reader R, to indicate its position. The whole instrument is so supported on a stand as that the plane of the circle can be brought into the plane of the angle which is to be measured. Having secured the telescope AB so that the reader attached to it may indicate zero, we bring both telescopes to point to one of the signals, and secure CD to the limb. Releasing AB, we now bring it to point to the second signal, taking care that CD still point to the former; the reader now indicates the angular distance between the two signals. AB being now secured to the limb, and CD released, we again bring both telescopes to the first object, secure CD in that position, release AB, and bring it once more to the second object, while CD points to the first; the reader now indicates the double of the angle.

By continuing this process, we can repeat the angle as often as we choose; then, dividing the ultimate reading by the number of the operations, we obtain the value of the angle with more precision than if we had made only one measurement, for the minute

Fig. 12.

Trigonometry. errors unavoidable in the graduation of the limb are subdivided. But no number of repetitions can give results true to a quantity less than that which is appreciable by the telescopes, or than what the stability of the instrument would warrant on a single observation, because each observation is accompanied by its own error, so that the sum of all the readings—that is, the ultimate reading of the instrument, includes as many of such errors as there are observations, so that the quotient must still be affected by the average error to which the instrument is liable.

The measurement of angles is most conveniently made by help of the sextant, or the reflecting circle. The construction of the sextant, and its use in astronomy, are given in the article NAVIGATION. The objects observed are there supposed to be remote, and the instrumental parallax has not been noticed; when the objects are near, this parallax becomes considerable, and we must therefore examine its source and the manner of correcting for it.

The inclination $OCI$ of the two mirrors is double of $SHP$ the angle which the two objects subtend, not at $C$ the centre of the instrument, but at a variable point $H$, so that, if $CP$ be joined, $CPH$ is the parallax or the difference between the true and the apparent angle, when the object $P$ is close at hand.

If the index $I$ be brought back a little behind the zero point $O$, until the mirror $AB$ be equally inclined to the lines $FC$, $CP$, the image of $P$ will be seen in the same direction with $P$; wherefore, if we first bring the direct and reflected images of that one of the signals which is to be looked at directly to agree, as when taking the index error of the instrument, and correct for that error in the usual way, we shall obtain the angle $CPS$ freed at once of parallax and of index error.

The sextant and octant, like all other fragmental instruments, are liable to this objection, that there are no means for ascertaining whether the actual axis of motion coincide with the centre of the graduation, or for correcting the error of centering; the observer has to rely entirely on the accuracy of the workmanship. By using an entire circle we obviate this inconvenience, and obtain also the advantage of repetition when great precision is required. The management of the reflecting circle does not differ, in any essential particular, from that of the sextant. There is in it this great advantage, that the left or the right hand object may be viewed directly.

Almost all the operations of the surveyor have reference to the positions which objects have on the surface of the earth; and hence, he is far oftener occupied in determining the directions of lines than in measuring their angles. These directions are, in general, indicated by referring them to the direction of the plummet, and to that of the meridian line; the first being obtained by processes purely mechanical, the second by the study of the motions of the stars. Frequently, indeed, the magnetic instead of the true north is used; but this is by no means a satisfactory practice, since the direction of the magnetised bar is subject to hourly, monthly, and secular variations, and is liable to great changes by change of position, so that the north shown by the compass at one station often differs by several degrees from that shown at another.

The simple plummet—that is, a fine thread with a small piece of lead attached so it affords the most obvious, and, for many purposes, a sufficient indication of the direction of gravity; but it is superseded almost for ordinary work, and altogether for accurate purposes, by the spirit-level, which is a glass tube nearly filled with alcohol. When this tube is laid almost horizontally, the bubble or empty space comes to the higher end, and its position thus indicates that end which may be the higher. The upper inside surface of the tube is made slightly concave in the direction of its length, the curvature, in fine levels, being given by grinding; but, in those for ordinary use, the bend which the glass takes while being drawn out is held sufficient. The air-bubble sweeping along the upper surface of the tube is thus the counterpart of the plummet bob; the one rests exactly under, the other exactly above, the centre of curvature. If the tube be ground to a curve of 28 feet 6 inches radius, an inclination of one minute causes the air-bubble to move through one-tenth part of an inch: such a level, which can be easily carried in the pocket, has as much precision as a plummet of 28 feet.

The tube is usually protected by a case, of which the under side is made accurately parallel to the surface at the middle, and divisions are marked along the glass to show the deviation from horizontality. In order, by help of such an instrument, to render a surface horizontal, or the normal to that surface vertical, it is

necessary to examine the level when placed in two directions inclined to each other; hence, in many cases, two cross levels are used.

The spherical spirit-level which indicates horizontality at once, consists of a circular disc of glass, having its under side ground concave; this disc is tightly fitted as the cover of a flat metal box, which is nearly filled with alcohol. The lower surface of the box, or the plane passing through the extremities of its three supports, is adjusted to be parallel to the spherical surface at the middle, and the air-bubble coming always to the highest point of the sphere, shows the deviation from horizontality, if there be any.

The directions of lines, in general, are ascertained by means of altitude-and-azimuth circles or theodolites. The essential parts of these instruments are as follows:—(1st), The tele-

scope $AB$ (Fig. 15), having cross wires in its field-bar, and turning in a vertical plane upon (2d) the horizontal axis $CD$. This horizontal axis carries (3d) a graduated vertical circle $E$, which shows the inclination of the telescope to the horizon, its angular distance from the zenith, or its nadir distance, according to the taste of the observer; the nadir distance is the most convenient. In the older theodolites, the telescope could only be raised or depressed 40 or 60 degrees; but in all good instruments the telescope may be turned completely round. The horizontal axis rests upon (4th) two pillars $CF$, $DE$, which are fitted with adjustments for rendering the axis truly horizontal. These again stand upon (5th) a plate or cross-bar $FHG$, fixed securely on the top of (6th) the vertical axis $KL$. This axis $KL$ works in a hollow axis $KL$ (7th), which ought to be perfectly concentric with $H$. By this arrangement the telescope $AB$ is free to be pointed in any required direction. There is attached to the outer axis $KL$, a horizontal or azimuth circle $M$ (8th), the divisions on which are read by verniers or micrometers attached to the cross-bar $FHG$; these show the bearing or azimuth of the telescope.

For the comfortable use of the instrument there are some secondary arrangements. The axis $KL$ is made to work into a frame provided with screws for rendering the axis truly vertical, there being a pair of cross-levels, or else a spherical spirit-level, fixed to the plate $FHG$ for assisting in this operation. By means of a strong clamp and tangent-screw the horizontal circle can be turned into and held in any required position. Since the reading of the horizontal angles would be vitiated by any displacement of the circle $M$, the better class of surveying circles carry a settler, that is, a telescope, $PQ$, fixed on the end of an axis $NO$, which turns in a frame that can be secured firmly in any required position on the axis $KL$. This settler, which ought to be quite as powerful as the principal telescope $AB$, is directed to any well-defined object, and then clamped firmly in its place; so that, if the circle $M$ be deranged, its position can be again put right by bringing the settler back to its signal by means of the lower tangent-screw. We have not room to detail the various adjustments of this useful instrument, and may only remark, that by reversing the position of the telescope, and taking the mean of the two sets of readings, the effect of any minute error in the horizontality of $CD$, or in the collimation of the telescope $AB$, is neutralised.

It is usual to reckon the bearings of objects from the north line, the degrees being numbered from north towards the east, so that a signal due east is said to bear 90°, one due south 180°, and one due west 270°; and therefore it is properly the first business of the surveyor to determine the true north.

For this purpose the most convenient proceeding is to establish a station for the theodolite near to head-quarters, and commanding a view of the district to be surveyed. At some distance from this a signal is to be set up in such a position that it may be illuminated by a lamp at night. Having then roughly estimated the bearing of that signal, we set the vernier to read that assumed bearing on the horizontal circle $M$, turn the whole instrument round until the telescope $AB$ points to the signal, secure the outer axis $KL$ in its place, and then bring the settler $PQ$ to the same, or to some other well-defined signal capable of being illuminated at night. The instrument is now in position for showing bearings as from the assumed meridian.

The latitude of the plane has now to be ascertained by observing the least or greatest zenith distance of a star (see ASTRONOMY, PRACTICAL, prob. ix, art 2, and NAVIGATION, chap. iii, sect. 1), preferring those stars which pass near to the zenith, and using observations with the face of the instrument alternately east and west. This done, the readiest process for obtaining the error of the assumed meridian is to take the extreme east or west bearing of some star which passes between the pole and the zenith, those which are near the pole being preferred; the true bearing of the star when in this position is easily computed, and the difference between this and the assumed bearing gives the required error.

Thus, in latitude $56^\{\backslash circ\}\; 23\text{'}$ north, the assumed bearing of a signal being $283^\{\backslash circ\}\; 20\text{'}$, the greatest easting of the star $\backslash beta$ Cephei was observed to be $24^\{\backslash circ\}\; 41\text{'}$ on March 11, 1860. On consulting the Nautical Almanac, we find that the north declination of the star at that date was $69^\{\backslash circ\}\; 56\text{'}\; 39\text{'}\text{'}$, whence

Having now obtained the true bearing of the signal, the position of the theodolite can be corrected, and the bearings of those station-points which are within sight determined.

In surveys of limited extent, embracing, say, only a few square miles, we may regard the surface of the earth as flat, and the meridian lines passing through the different points as all parallel to each other. In this case it is easy to transfer the theodolite to another station; for, if the bearing of B from A be $297^\{\backslash circ\}\; 05\text{'}$, that of A from B must be $117^\{\backslash circ\}\; 05\text{'}$, the difference being $180^\{\backslash circ\}$. Hence, if the theodolite be carried to B, and so placed as that the bearing of A be $117^\{\backslash circ\}\; 05\text{'}$, its zero, or north line, will be parallel to what it was at A; in this way the instrument may be carried from station to station, until the directions of all the principal lines of the survey be ascertained.

When the ground is uneven, it is necessary to notice the inclinations of the lines: this is conveniently done by painting the signal-staves alternately black and white at each link or foot, and by directing the telescope to the division corresponding to the height of the instrument. On the common theodolite, the zero on the vertical circle shows the telescope to be horizontal, and we must distinguish elevation (+) from depression (-).

The conductor of the survey, after having obtained a general idea of the configuration of the district, proceeds to arrange the stations, choosing these so that the lines joining them may pass near to the boundaries of the fields, at the same time having regard to their being conspicuous. He then ascertains, in the manner above described, the directions of these as seen from each other, entering the observations in his Theodolite-Book; and afterwards he proceeds to measure such of their distances as may be required, and the offsets or deviations of the actual boundaries from the measured lines, recording these measurements in his Chain-Book.

The theodolite-book is conveniently divided into five columns, the first for the name or number of the station at which the instrument is planted; the second for the actual readings on the horizontal limb; the third for the readings on the vertical arc; the fourth for the measured distances (to be extracted from the chain-book); and the fifth for the name or number of the signal. The following extract will sufficiently exemplify this matter:—

The page of the chain-book is divided into three columns, the middle one to receive the distances measured along the traverse lines, and those to the right and left for offsets to either side. The beginning of each page is at the bottom, and the writing proceeds upwards, so as to imitate the progress on the field; and at the commencement of each line its bearing, as extracted from the theodolite-book, is given. The following is the chain-work for the field ABCDEFG.

In making the plan, many surveyors use the protractor, and imitate on paper the operations which have been performed in the field; this plan, again, they re-measure for the area. This proceeding is liable to the very serious objection that the errors on paper far exceed those to which the field operations are liable. It is easier and preferable to compute the co-ordinates of the stations, and by help of these to lay down the plan and compute the areas. If the ordinates $x$ be directed northwards, the $y$'s eastwards, and the $z$'s towards the zenith, the difference between the ordinates of one station, as A, and those of another station, as B, are given by the equations,—

When the co-ordinates of any one of the stations are either known or assumed, those of each of the other stations can be easily deduced by means of these formulae, using either the ordinary trigonometrical tables or the traverse tables. Thus, if we assume for the station A, $x\_A\; =\; 500$, $y\_A\; =\; 100$, $z\_A\; =\; 150$, we have,

Having proceeded from A successively to B, C, D, E, F, and G, we return from G to A. If all the operations have been absolutely exact, the co-ordinates of A thus obtained should have agreed exactly with those from which we first set out. Actually there are small errors—viz., $-4$ in the direction x, $-1.1$ in the direction y, and $+5$ in the direction z. These may naturally be attributed to the unavoidable minute errors in the measurements; and the close coincidence shows that no serious error has been committed. The positions of the several points may be taken as under, rejecting the fractional parts of a link:—

From these results, it is easy to mark on the plane the positions of the stations, and then to lay down the contour of the figure from the chain-book.

The area of the polygon is readily computed from the formula—
$2\; \backslash text\{\; Area\}\; =\; x\_A(y\_B\; -\; y\_A)\; +\; x\_B(y\_C\; -\; y\_B)\; +\; \backslash dots$

This formula is easily kept in mind: we proceed round the polygon from corner to corner, and multiply the x or latitude of the point at which we are by the y or longitude of the point before us minus the y of the point behind us. The aggregate of these products, attention being paid to the algebraic signs, is the double of the area.

The computation for the preceding example stands thus:—

The areas included between the traverse lines and the actual boundaries of the field have now to be computed and added to or subtracted from the area of the polygon according to their positions; the subjoined computations of the offsets from A to B and from G to A exemplify sufficiently well the manner of carrying on the work. Strictly, the results have to be multiplied by the cosines of the inclinations of the respective traverse lines, in order to give the areas of the projections on a horizontal plane.

A to B.

The exact area of the figure is thus:—

or 7.610 acres.

When a surveyor has carried his operations with the theodolite over a considerable distance, he is naturally led to inquire whether the instrument now show a north line exactly parallel to what it did at the outset. In order to avoid uncertainty in this respect, he seeks to reduce as far as possible the number of transferences, or even uses a more powerful instrument for connecting the bearings of the distant stations. He can readily obtain a verification of his work by appealing again to the astronomical method; but here he meets with this difficulty, that the true meridians of two places are not parallel to each other, the north lines converging on the north side and diverging on the south side of the equator.

In order to compute the convergence of two meridian lines, we may observe that these would meet each other on the prolongation of the earth's axis; hence the easting or westing divided by the cotangent of the terrestrial latitude must measure the convergence: now, one minute of longitude on the earth's equator measures 6086 feet or 9221 links; hence, if we divide the difference between the y's of two stations by 6086 or by 9221, according as our measurements are in feet or in links, and multiply the quotient by the tangent of the latitude, we obtain the convergence of the meridians in minutes. Thus, in our example of a very small survey made in north latitude $56^\backslash circ$, the convergence of the meridians at the extreme east and west stations is computed as under:—

which gives $0.2469$ or $14.4$, about a quarter of a minute.

When the survey extends over many square miles, we can no longer neglect the curvature of the earth in any one of the operations; and even the oblateness must be taken into account. (E. S.)

Trikala TRIKHALA, or TRICALA (anc. Tricca), a town of European Turkey, Thessaly, 37 miles W.N.W. of Larissa. It covers a large space of ground, and contains seven mosques, ten Greek churches, two synagogues, and large bazaars. Blankets, coarse woollen and cotton cloth, are made; and there are dye-works for Turkish yarn. A considerable transit trade is carried on. Pop. 12,000.

TRIM, a town of Ireland, capital of the county of Meath, on the left bank of the Boyne, 25 miles N.W. of Dublin. It is an ancient town; and had formerly walls, parts of which still remain. There are also the ruins of a large and strong castle, and of an old abbey. It has a parish church, Roman Catholic chapel, schools, court-house, jail, workhouse, &c. Trim has some trade in rural produce. Pop. 1905.

TRINCOMALIE, a town in Ceylon, with a harbour the most capacious and secure in the Indian seas. It is situated in the north-east portion of the island, Lat. 8. 35. 38. N., and Long. 81. 16. 37. E., distant by the road 182 miles from Colombo, 130 from Jaffna, and about 113 from Kandy, the central capital. The coast in the immediate vicinity is bold and rocky, rising into wooded acclivities, with forests of valuable timber-trees, amongst which satinwood and ebony are abundant. The shore near the villages, wherever sand has been heaped up by the currents and rivers, is planted with palms, chiefly the cocoa-nut and palmyra (Borassus flabelliformis).

The entrance to the Bay of Trincomalee is by a passage from four to five miles wide; but such is the force of the current that sets to the southward during the N.E. monsoon, that sailing vessels are frequently carried past it. On freeing the channel the inlet expands, headlands and islands dividing the inner harbour to the north from Great Bay to the south. The latter receives the waters of the Mahawelli-ganga, the Kottiar, and two other rivers; it is about 5 miles in width, with a similar length, and its depth in some places is upwards of 70 fathoms. To the west of Great Bay, and communicating with it by a passage navigable only by boats, is the lagoon of Tamblegam, a shallow lake about 20 miles in circumference, which, according to a native tradition, was at no remote period a broad expanse of rice lands, irrigated by a canal from the enormous tank at Kandelay, 24 miles to the westward. The tank having partially fallen to ruin; and the waters issuing in a torrent, converted their ordinary outlet into an impetuous river, which, overflowing the plain below, burst open an entrance for the sea, that, once admitted, has ever since continued to hold possession. An examination of the locality confirms to some extent the possible truth of this tradition. The remains of the great tank are still in fine preservation, and could be readily restored; but the waters escaping from the broken bund, although partially applied to cultivation, are to some extent lost in the lagoon of Tamblegam. The lake abounds in fish, and produces in perfection the thin transparent oyster (Placusa placenta), whose clear white shells are used in China and elsewhere as a substitute for window-glass. They are collected annually for the sake of the pearls they produce. These pearls are exported to the coast of India, there to be burned into a species of lime, which the more luxurious princes affect to chew with their betel. So prolific are the mollusca of the placusa, that the number of shells taken by the licensed renter in the three years prior to 1858 could not have been less than 18,000,000.

The inner harbour of Trincomalee is land-locked in every direction, with safe anchorage and deep water close to the principal wharves. It is protected on the east side by a rocky peninsula between 3 and 4 miles long, whose southern extremity rises into two elevated ridges, one of which, above Osterberg Point, sustains a fort for the defence of the inner bay. The entrance to the latter is little more than a quarter of a mile wide. These cliffs, and the basalt of which they are partially composed, exhibit traces of a volcanic origin, a conjecture which is sustained by the existence of a hot spring at Kannea, a few miles to the north-west. Some shoals and sunken rocks, well known to the local pilots, render the passage somewhat hazardous without their assistance. Within the area of the great harbour it covers a space upwards of 3 miles in length from north to south, with a breadth somewhat greater, irrespective of the coves and indentures which diversify its outline. It contains several wooded islands, and it is surrounded by hills covered with forests to their summits.

On the east side of the peninsula, which protects the inner harbour, the outline of the coast forms another port, fronting the Bay of Bengal, known as Back Bay, which affords safe holding ground

and deep water close to the beach. This is resorted to by shipping Trincomalee during the S.W. monsoon, when the wind is off the shore. On a promontory at its southern extremity, which is surmounted by Fort Frederick, a light is displayed at a height of 205 feet above the sea, and serves to guide vessels clear of the rocks off Pigeon Island to the north, as well as of the shoals off Foul Point, at the south side of the passage leading to the bay. From the 25th October, when the N.E. monsoon begins to blow, till the middle of March, the anchorage in Back Bay is unsafe, owing to the heavy swell which it occasions.

Fort Frederick is commanded from the adjacent heights; and being situated 3 miles to the northward of the dockyard and the mouth of the inner harbour, it protects only the outer anchorage, and is available solely as a point d'appui. Notwithstanding their extent, the military works of Trincomalee are utterly incommensurate with the importance of the position, and would be found ineffectual for its protection in the event of attack. The town is built across the neck of the promontory above referred to, and extends to the sea on either side. It had formerly a bad reputation from fever and other ailments incidental to damp and malaria, but the clearance of the jungle close to the suburbs, and the draining of the adjacent lands previously submerged by the decay of works for irrigation, have contributed to remove this reproach, and Trincomalee, at the present day, is no less healthy than any other portion of the maritime provinces. Its climate resembles that of the Indian peninsula south of Madras. With moist winds and plentiful dew light showers are frequent, and the rain throughout the year does not exceed 40 inches. The temperature of this part of Ceylon follows the course of the sun, and ranges from a minimum of 70° in December and January to a maximum of 94° in May and June; but the heat is rendered tolerable at all seasons by the steadiness of the land and sea breezes. The mean is about 81.4. A meteorological record for the year 1854 exhibits the following results:—

Trincomalee, which is still called by the Tamils, who form the preponderating body of the population, Tiru-konatha-malie, "the sacred hill of Konatha," appears to have been one of the earliest settlements of the Malabar race in Ceylon, who, at a period which has escaped historic record, erected on the cliff, now crowned by Fort Frederick, a temple dedicated to Konatha or Konasir. This edifice was still in existence on the arrival of the Portuguese, in the early part of the sixteenth century. They desecrated and destroyed it in 1622, using the materials for the construction of the fort; but the pagoda is remembered as the "Temple of a Thousand Columns," and near its site is a cliff still designated the Sausay Rock, a term expressive of its sanctity. Twice a week a procession, attended by devotees, who bring offerings of fruits and flowers, repairs at sunset to the spot where the rock projects above the ocean; a series of ceremonies are performed, including the mysterious breaking of a cocoa-nut against the cliff; and the officiating Brahman concludes his invocation by elevating a brazen censer above his head, filled with inflammable materials, the light of which, as they burn, is reflected far over the sea.

The Portuguese attached no importance to the possession of Trincomalee, which lay remote from their trading settlements on the opposite side of Ceylon; and their attention was only directed towards it by their alarm for its falling into the hands of Holland. Their imperfect fortifications did not avert this catastrophe; the Dutch took possession of it in 1639, and under their authority it remained till temporarily occupied by the English under Sir Hector Munro in 1782. The fort was, however, surprised by Admiral Suffrein and a French fleet immediately afterwards, and restored to Holland in the following year. At length, in 1795, an expedition fitted out from Madras arrived, and, after a siege of three weeks, reduced the fort and the town. Trincomalee, with all other