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ular an indefinite length; take 40 in the dividers, and setting one foot in A, wherever the other foot strikes the perpendicular will be the point C.

NOTE. When the triangle is constructed, the angles may be measured by a protractor, or by a scale of chords. Fig. 33.

PROBLEM XII. To make a right angled triangle, the two legs being given. Fig. 33.

38

46

Suppose the leg AB 38, and the leg BC 46.

Draw the leg AB in length 38; from B erect a perpendic ular to C in length 46; and draw a line from A to C.

PROBLEM XIII. To

Fig. 34.

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Suppose the side BC 98; the angle at B 45° 15', the angle at D 108o 30', consequently the other angle 26° 15'.

Draw the side BC in length 98; ; on the point B make an angle of 45o 15'; on the point C make an angle of 26° 15′, and draw the lines BD and CD.

PROBLEM XIV. To make an oblique angled triangle, two sides and an angle opposite to one of them being given. Fig. 35.

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Suppose the side BC 160, the side BD 79, and the angle at C 29° 9'.

Draw the side BC in length 160; at C make an angle of 29° 9', and draw an indefinite line through where the degrees cut the arc; take 79 in the dividers, and with one foot in B lay the other on the line CD; the point D will be the other

Fig. 36.

PROBLEM XV. To make an ob- D lique angled triangle, two sides and their contained angle being given. Fig. 36.

209

Suppose the side BC 109, the side BD 76, and the angle at B 101° 30'.

Draw the side BC in length 109; at B make an angle of 101° 30′, and draw the side BD in length 76; draw a line from D to C and it is done.

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Draw the line AB the length of the proposed Square; from Berect a perpendicular to C and make it of the same length as AB; from A and C, with the same distance in the dividers, describe arcs intersecting each other at D, and draw the lines AD and DC.

PROBLEM XVII. To make a rectangle. Fig. 38.

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Draw the line AB equal to the longest side of the rec tangle; on B erect a perpendicular the length of the shortest side to C; from C, with the longest side, and from A, with the shortest side, describe arcs intersecting each other at D,

PROBLEM XVIII. To describe a circle which shall pass through any three given points, not lying in a right line, as A, B, D. Fig. 39.

B

D

Fig. 39.

Draw lines from A to B and from B to D; bisect those lines by PROBLEM II. and the point where the bisecting lines intersect each other, as at C, will be the centre of the circle.

PROBLEM XIX. To find the centre of a circle.

By the last PROBLEM it is plain, that if three points be any where taken in the given circle's periphery, the centre of the circle may be found as there taught.

Directions for constructing irregular figures of four or more sides may be found in the following treatise on SURVEYING.

TRIGONOMETRY.

TRIGONOMETRY is that part of practical GEOMETRY by which the sides and angles of triangles are measured; whereby three things being given, either all sides, or sides and angles, a fourth may be found; either by measuring with a scale and dividers, according to the PROBLEMS IN GEOMETRY, or more accurately by calculation with logarithms, or with natu

ral sines.

TRIGONOMETRY is divided into two parts, rectangular and oblique-angular.

PART I.

RECTANGULAR TRIGONOMETRY.

This is founded on the following methods of applying a circle to a triangle.

PROPOSITION I. In every right angled triangle, as ABC, Fig. 40, it is plain from Fig. 7, compared with the Geometrical definitions to which that Figure refers, that if the hypothenuse A AČ be made radius, and with it an arc of a circle be described from each end, BC will be the sine of the angle at A, and AB the sine of the angle at C; that is, the legs will be sines of their opposite angles.

PROPOSITION II. If one leg, AB, Fig. 41, be made radius, and with it on the point A an arc be described, then BC, the other leg, will be the tangent, and AC the secant of the angle at A ; and if BC be made radius, and an arc be described with it on the point C, then AB will be the tangent and AC the se-A cant of the angle at C; that is, if one leg

Fig. 40.

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be made radius the other leg will be a tangent of its opposite

Thus, as different sides are made radius, the other sides acquire different names, which are either sines, tangents or secants.

As the sides and angles of triangles bear a certain proportion to each other, two sides and one angle, or one side and two angles being given, the other sides or angles may be found by instituting proportions, according to the following rules.

RULE I. To find a side, either of the sides may be made radius, then institute the following proportion:

AS THE NAME OF THE SIDE GIVEN, (which will be either radius, sine, tangent or secant ;)

IS TO THE LENGTH OF THE SIDE GIVEN ;

SO IS THE NAME OF THE SIDE REQUIRED, (which also will be either radius, sine, tangent or secant ;)

TO THE LENGTH OF THE SIDE REQUIRED.

RULE II. To find an angle one of the given sides must be made radius, then institute the following proportion : AS THE LENGTH OF THE GIVEN SIDE MADE RADIUS;

IS TO ITS NAME, (that is radius ;)

So IS THE LENGTH OF THE OTHER GIVEN SIDE;

TO ITS NAME, (which will be either sine, tangent or secant.)

Having instituted the proportion, look for the corresponding logarithms, in the logarithms of numbers for the length of the sides; and in the table of artificial sines, and tangents, and for the logarithmic sine, tangent or secant.

ADD TOGETHER THE LOGARITHMS OF THE SECOND AND THIRD TERMS, AND FROM THEIR SUM SUBTRACT THE LOGARITHM OF THE FIRST TERM: THE REMAINDER WILL BE THE LOGARITHM OF THE FOURTH TERM, WHICH SEEK IN THE TABLES, AND FIND ITS CORRESPONDING NUMBER, OR DEGREES AND MINUtes.

See the introduction to the table of logarithms; which should be attentively studied by the learner before he proceeds any further.

NOTE. The logarithm for radius is always 10, which is the logarithmic sine of 90°, and the logarithmic tangent of

45.

The preceding PROPOSITIONS and RULES being duly attended to, the solution of the following CASES of Rectangular Trigonometry will be easy.

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