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cotangent, and cosecant form a right-angled triangle CDL. All these right-angled triangles are similar to each other. 1046. The sine, tangent, or secant of an angle is the sine, tangent, or secant of the are by which the angle is ‘measured, or of the degrees, &c. in the same arc or angle. The method of constructing the scales of chords, sines, tangents, and secants engraved on mathematical instruments is shown in the annexed figure. 1047. A trigonometrical canon (fig. 396.) is a table wherein is given the length of the sine, tangent, and secant to every degree and minute of the quadrant, compared with the radius, which is expressed by unity or 1 with any number of ciphers. The logarithms, moreover, of these sines, tangents, and secants, are tabulated, so that trigonometrial calculations are performed by only addition and subtraction. Tables of this sort are published separately, and we suppose the reader to be provided with them. 1048. PRoBLEM I. To compute the natural sine and cosine of a given arc. The semiperiphery of a circle whose radius is 1 is known to be 3.141592653589793, &c.; we have then the following proportion: –

As the number of degrees or minutes in the semicircle
Is to the degrees or minutes in the proposed arc,
So is 3.14159265, &c. to the length of the said arc.

Now the length of the arc being denoted by the letter a, and its sine and cosine by s and c, these two will be expressed by the two following series, viz. –

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Example 1. Let it be required to find the sine and cosine of one minute. The number of minutes in 180 degrees being 10800, it will be, first, as 10800 : 1:3-14159265, &c. : (MOO290888208665 = the length of an arc of one minute. Hence, in this case,— - a = 0002908882 and ja3 = 000000000004 The difference is s = 0.002908882, the sine of one minute. Also from 1. take a*=0.0000000423079, &c. leaves c = .9999999.577, the cosine of one minute. Example 2. For the sine and cosine of 5 degrees : – Here 180°: 5°:.3.14159265, &c. : 08726646=a, the length of 5 degrees. Hence a = -08726646 —las = 00011076 + Tina” = 00000004

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These collected give c = .996.19470, the cosine of 5 degrees.

In the same way we find the sines and cosines of other arcs may be computed. The greater the arc the slower the series will converge; so that more terms must be taken to make the calculation exact. Having, however, the sine, the cosine may be found from it

by the property of the right-angled triangle CBF, viz. the cosine CF= x/CB7-BF,

or c = x/1-52. There are other methods of constructing tables, but we think it unnecessary to mention them; our sole object being here merely to give a notion of the mode by which such tables are formed.

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1049. Paoh. II. To compute the tangents and secants. 1Iaving, by the foregoing problem, found the sines and cosines, the tangents and secants are easily found from the principles of similar triangles. In the arc AB (fig. 395.), where BF is the sine, CF or BK the cosine, A H the tangent, CII the secant, DL the cotangent, and CL the cosecant, the radius being CA or CB or CD; the three similar triangles CFB, CAH, CDL, give the following proportions: — I. CF : FB:: CA : AH, by which we find that the tangent is a fourth proportional to the cosine, sine, and radius. II. CF : CB:: CA : CH, by which we find that the secant is a third proportional to the cosine and radius. III. BF : FC:: CD : DL, by which we find that the cotangent is a fourth proportional to the sine, cosine, and radius. IV. B F : BC:: CD : CL, by which we find that the cosecant is a third proportional to the sine and radius. Observation 1. There are therefore three methods of resolving triangles, or the cases of trigonometry; viz. geometrical construction, arithmetical computation, and instrumental operation. The method of carrying out the first and the last does not need explanation: the method is obvious. The second method, from its superior accuracy in practice, is that whereof we propose to treat in this place. Observation 2. Every triangle has six parts, viz. three sides and three angles. And in all cases of trigonometry, three parts must be given to find the other three. And of the three parts so given, one at least must be a side; because, with the same angles, the sides may be greater or less in any proportion. observation 3. All the cases in trigonometry are comprised in three varieties only; w1z. 1st. When a side and its opposite angle are given. 2d. When two sides and the contained angle are given. 3d. When the three sides are given. More than these three varieties there cannot possibly be; and for each of them we shall give a separate theorem. 1050. THEoREM I. When a side and its opposite angle are two of the given parts. Then — the sides of the triangle have the same proportion to each other as the sines of their opposite angles have. That is,

As any one side
Is to the sine of its opposite angle,
So is any other side
To the sine of its opposite angle.

For let ABC (fig. 397.) be the proposed triangle, having AB the greatest side, and BC the least. Take AD as a radius equal to BC, and let C fall the perpendiculars DE, CF, which will evidently be the sines of the angles A and B, to the radius AD or BC. Now the triangles ADE, ACF are equiangular; they therefore have their like sides proportional, namely, AC : CF::AD or BC : DE, that is, the sine AC is to A the sine of its opposite angle B as the side BC is to the sine of its opposite angle A. Note 1. In practice, when an angle is sought, the proportion is to be begun with a side opposite a given angle; and to find a side, we must begin with the angle opposite the given side. Note 2. By the above rule, an angle, when found, is ambiguous; that is, it is not certain whether it be acute or obtuse, unless it come out a right angle, or its magnitude be such as to remove the ambiguity; inasmuch as the sine answers to two angles, which are supplements to each other; and hence the geometrical construction forms two triangles with the same parts, as in an example which will follow: and if there be no restriction or limitation included in the question, either result may be adopted. The degrees in a table answering to the sine is the acute angle; but if the angle be obtuse, the degrees must be subtracted from 180 degrees, and the remainder will be the obtuse angle. When a given angle is obtuse, or is one of 90 degrees, no ambiguity can occur, because neither of the other angles can then be obtuse, and the geometrical construction will only form one triangle.

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Example 1. In the plane triangle ABC,
Let A B be 345 feet,
B.C 232 feet,
Z A 37 - 20':
Required the other parts.
First, to the angles at C and B (fig. 398.)

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It is to be observed here that the second and third logarithms are added (that is, the numbers are multiplied), and from the sum the first logarithm is subtracted (that is, division by the first number), which leaves the remainder 9.955127, which, by the table of sines, is found to be that of the angle 115° 36', or 64° 24'.

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1051. THEoREM II. When two sides and their contained angle are given. The given angle is first to be subtracted from 180° or two right angles, and the remainder will be the sum of the other two angles. Divide this remainder by 2, which will give the half sum of the said unknown angles; and using the following ratio, we have – As the sum of the two given sides Is to their difference, So is the tangent of half the sum of their opposite angles To the tangent of half the difference of the same angles. Now the half sum of any two quantities increased by their half difference gives the greater, and diminished by it gives the less. If therefore, the half difference of the angles above found be added to their half sum, it will give the greater angle, and subtracting it will leave the lesser angle. All the angles thus become known, and the unknown side is then found by the former theorem. For let ABC (fig. 399.) be the proposed triangle, having the two given sides AC, BC, including the given angle C. With the centre C and radius E CA, the less of these two sides, describe a semicircle, meeting the other side of BC produced in E, and the unknown side A B in G. Join AE, CG, and draw DF parallel to A.E. Now BE is the sum of the given sides AC, CB, or of EC, CB; and BD is the difference of these given sides. The external angle ACE is equal to the sum of the two internal or given angles CAB, CBA; but the angle ADE at the circumference is equal Fig. 5.19. to half the angle ACE at the centre; wherefore the same angle ADE is equal to half the sum of the given angles CAB, CBA. Also the external angle AGC of the triangle BGC is equal to the sum of the two internal angles GCB, GBC, or the angle GCB is equal to the difference of the two angles AGC, GBC; but the angle CAB is equal to the said angle AGC, these being opposite to the equal sides AC, CG, and the angle DAB at the circumference is equal to half the angle DCG at the centre. Therefore the angle DAB is equal to half the difference of the two given angles CAB, CBA, of which it has been shown that ADE or CDA is the half sum.

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Now the angle DAE in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD: therefore AE is the tangent of CDA the half sum; and DF, the tangent of DAB, the half difference of the angles to the same radius AD, by the definition of a tangent. But the tangents A E, DF being parallel, it will be as BE : BD:: A E : DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles to the tangent of half their difference.

It is to be observed, that in the third term of the proportion the cotangent of half the given angle may be used instead of the tangent of the half sum of the unknown angles.

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1052. THEoREM ll I. When the three sides of a triangle are given. Let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles, the proportion will be— As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base.

Then take half the difference of these segments, and add it to the half sum, or the half base, for the greater segment; and for the lesser segment subtract it. Thus, in each of the two right-angled triangles there will be known two sides and the angle opposite to one of them, whence, by the first theorem, the other angles will be found. For the rectangle under the sum and difference of the two sides is equal to the rectangle under the sum and difference of the two segments. Therefore, forming the sides of these rectangles into a proportion, their sums and differences will be found proportional. Example. In the plane triangle ABC (fig. 401.), C Let AB =345 ft. AC =232 ft.

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Sives the whole angle ACB = 115 36 So that the three angles are as follow, viz. & A 27 4'; & B 37°20'; Z C 115° 36. 1053. THEoREM IV. If the triangle be right-angled, any unknown part may be found by tha following proportion:

As radius
Is to either leg of the triangle,
So is tangent of its adjacent angle
To the other leg;
And so is secant of the same angle
To the hypothen use.

For AB being the given leg in the right-angled triangle ABC, from the centre A with any assumed radius A D describe an arc DE, and draw t DF perpendicular to AB, or parallel to BC. Now, from the definitions, / DF is the tangent and A F the secant of the arc DE, or of the angle A, * which is measured by that arc to the radius AD. Then, because of the E/: |

D b.

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parallels BC, DF, we have AD : A B:: D F : BC, and ::AF : AC, which

is the same as the theorem expresses in words.
Note. Radius is equal to the sine of 90, or the tangent of 45°, and is A

expressed by 1 in a table of natural sines, or by 10 in logarithmic sines.
Example 1. In the right-angled triangle ABC,

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Note. There is another mode for right-angled triangles, which is as follows: – ABC being such a triangle, make a leg AB radius; or, in other words, from the centre A with distance AB describe an arc B F. It is evident that the other C leg BC will represent the tangent and the hypothenuse AC the se- P cant of the arc B F or of the angle A. In like manner, if BC be taken for radius, the other leg A B repre- G sents the tangent, and the hypothenuse AC the secant of the arc BG or angle C. -----If the hypothenuse be made radius, then each leg will represent * ; B to the sine of its opposite angle; namely, the leg AB the sine of the E. are A E or angle C, and the leg BC the sine of the arc CD or angle A. Then the general rule for all such cases is, that the sides of the triangle bear to each other the same proportion as the parts which they represent. This method is called making every side radius. 1054. If two sides of a right-angled triangle are given to find the third side, that may be found 'y the property of the squares of the sides (Geom. Prop. 32.; viz. That the square of the hypothemuse or longest side is equal to both the squares of the two other sides together). Thus, if the longest side be sought, it is equal to the square root of the sum of the squares of the two shorter sides; and to find one of the shorter sides, subtract one square from the other, and extract the square root of the remainder. 1055. The application of the foregoing theorems in the cases of measuring heights and dist inces will be obvious. It is, however, to be observed, that where we have to find the length of inaccessible lines, we must employ a line or base which can be measured, and, by means of angles, which will be furnished by the use of instruments, calculate the lengths of the ot!:er lines.

Fig. 403.

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