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XXVIII All triangles that are not right angled, whether they are acute or obtuse, are, in general terms, called OBLIQUE ANGLED TRIANGLES, without any other distinction.

Also, the three sides of a triangle are frequently distinguished by giving to one of them the name of base, in which case the other two are called the two sides, and the angular point opposite to the base is called the vertex (or summit). In an isosceles triangle the vertex is the angular point between the two equal sides, and the base the side opposite to it.

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400

XXIX The SINE of an arc is a line drawn from one extremity of the arc, perpendicular to a diameter drawn to the other extremity, and is equal to half the chord of double the arc : thus D E is the sine of the arc D B, and is equal to half the chord D I, which is the chord of the arc D B 1—the double of the arc D B.

500

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The VERSED SINE of an arc is that part of a diameter contained between the sine and the arc; thus E B is the versed sine of the arc D B; and E L is sometimes called the SUVERSED SINE.

The TANGENT of an arc is a line drawn perpendicular to the end of a diameter passing through one extremity of the arc, and continued till it meets a line drawn from the centre through the other end of the arc; thus A B is the tangent of the arc D B.

The SECANT of an arc is a line drawn from the centre of the circle through one end of the arc, till it meets the tangent drawn from the other end ; thus C A is the secant of the arc D B.

The COMPLEMENT of an arc is what it wants of a right angle, or 90 degrees ; thus G D is the complement of D B, or D B of D.G.

The SUPPLEMENT of an arc is what it wants of two right angles, or 180 degrees; thus L D is the supplement of D B, or D B of L D.

The Co-SINE, CO-TANGENT, CO-SECANT, and CO-VERSED SINE of an arc, are the sine, tangent, secant, and versed sine of the complement of that arc; Co. being a contraction of the word complement : thus D F is the cosine, G H the cotangent, C.H the cosecant, and G F the coversed sine of the arc D B; being the sine, tangent, etc., of the arc D G, the complement of the arc D B.

The sine, tangent, and secant of an arc, as of D B, is likewise the sine, tangent, and secant of the supplement of that arc, as of L D.

An angle being measured by an arc of a circle (see Def. XIV.), the sine, tangent, etc., of an arc is the sine, tangent, etc., of the angle which is measured by the arc, or of the degrees and minutes, etc., that the arc contains ; hence, supposing the arc D B, which measures the angle D C B, to contain 50 degrees, the lines D E, AB, AC, and E B, will be respectively the sine, tangent, secant, and versed sine of the angle A C B, or of 50 degrees; and consequently the cosine, cotangent, cosecant, and coversed sine, of the angle G C D, or of 40 degrees, which is the complement of 50 degrees.

GEOMETRY

PROBLEM I

To divide a given Line A B into two equal Parts

Take any extent in the compasses greater than half the line A B, and with one foot in B describe an arc; with the same radius, and one foot in A, describe an arc cutting the former in C and D; through C and D draw a line; and this line will divide the given line A B into two equal parts at the point E.

In this manner any arc of a circle may be divided into two equal parts

E

A

PROBLEM II

From a given Point C, in a given Line A B, to raise a Perpendicular

CASE ist. When the given point C is near the middle of the line A B.

With one foot of the compasses in C, at any distance, draw an arc cutting the line A B in D and E; from the points D and E, with any distance greater than CE or CD, describe two arcs cutting each other in F; through the points F and C draw the line I C, and it will be

perpendicular to the given line A B.

А

CASE 2nd. When the given point C is at, or near the end of the line A B.

Take any point out of the line, as D, and with the distance D C describe a circle, cutting the line A B in E and C; through the centre D and the point E draw the line E F, cutting the circle in F; then a line drawn through F and C will be the perpendicular required.

Or thus : Describe the arc D E at any distance from C; and with one foot of the compasses in D, with the same extent, describe an arc cutting the arc D E in E; from this point, keeping the same extent in the compasses, draw the arc G; through D and E draw the line DG, cutting the arc in G; then draw a line through G and C, and it will be the perpendicular required.

A

PROBLEM III

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From a given Point C to let fall a Perpendicular on a given Line A B

CASE ist. When the point C is nearly opposite the middle of the line A B.

With one foot of the compasses in C, describe an arc cutting the line A B in D and E; from these points, at any distance, describe two arcs cutting each other in F; through the points C and F draw a line, and it will be perpendicular to the given line A B.

CASE 2nd. When the given point C is nearly opposite to the end of the line AB.

Place one foot of the compasses in any part of the given line, as at A, and with the distance AC describe the arc C E; then from any other part of the given line nearly under the point C, as at D, with the distance D C describe a small arc cutting the arc C E in E; then a line drawn through the points C and E will be perpendicular to the line AB.

А

PROBLEM IV

To draw a Line parallel to a given Line A B. CASE ist. When the parallel line is to pass through a given point D.

Take the nearest distance between the given point D and the line A B; with that distance set one foot of the compasses on any part of the line A B, as at C, and describe the arc E; from the point D draw a line so as just to touch the arc E without cutting it; then that line will be parallel to the given line A B, and pass through the given point D.

A

CASE 2nd. When the parallel line is to be at a given distance from the line A B.

With the given distance in the compasses, describe two arcs D and E, from any two points, as F and G in the given line ; then a line D E drawn just touching the two arcs without cutting them, will be parallel to the given line A B.

PROBLEM V

At a given Point in a Line to make an Angle equal to a given Angle

The given angle C B A; and D is the point in the line D F. With one foot of the compasses in B describe the arc A C: with the same extent in the compasses, place one foot in D and describe the arc HE; then take

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the distance A C and apply it to the arc H E from H to G, and through the points D and G draw the line D K; the angle G D H will then be equal to the angle CBA, as required.

PROBLEM VI

To divide a given Angle A B C into two Equal Parts From the angular point B, with any extent in the compasses, describe the arc D E; from D and E, with the same or any other extent, describe two arcs cutting each other in F; through the points B and F draw a line, and it will divide the angle into two equal parts.

In the same manner any given arc of a circle is bisected, when the centre of the circle is given.

PROBLEM VII

To divide a Circle A B C D into two, four, eight, sixteen, thirty-two, etc., Equal

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PROBLEM VIII
To describe the Circumference of a Circle through any three given

Points A B C, not situated in a straight Line
Draw lines joining A B and B C, and bisect
them by lines meeting in 0, as directed in
Problem I. ; then from 0, at the distance of
any one of the points, as 0 A, describe a circle,
and it will pass through the other points B
and C, as required.

In this manner the centre of a circle may be found; for, taking any three points in the circumference, and proceeding as directed above, the lines meeting at will give the centre required.

PROBLEM IX

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To divide a given Line A B into any proposed Number of Equal Parts

Let it be required to divide the line A B into seven equal parts ; from one end A, of the given line A B, draw a line A C, making any angle with A B, and from the other end B draw a line B D parallel to AC; on each of the lines A C, BD, beginning at A and B, set off as many equal parts as A B is to be divided into, viz. seven; then lines drawn from A to 7, 1 to 6, 2 to 5, 3 to 4, etc., will divide the given line into seven equal parts.

PROBLEM X

To construct Scales of Equal Parts The simplest scale of equal parts is made by drawing a straight line, and dividing it with a pair of compasses into as many primary divisions as convenient, which, if the line be of a definite length, may be done by Problem IX. ; subdivide one of these divisions decimally, i.e. into 10 equal parts; then each of the former may represent 10 units, as leagues, miles, etc., and in that case the latter will represent one of these units : or if the larger divisions be supposed to be 100, then the subdivisions will be tens, and so on.

There are frequently several of these scales drawn parallel to each other, of different lengths, on a flat rule (as Fig. 1, Plate I.); they are divided into as many equal parts as the length of the rule will admit; the numbers placed on the left hand showing how many parts in an inch each scale is divided into

But the most correct scale of equal parts is the DIAGONAL SCALE (Fig. 2, Plate I.), the larger divisions of which are commonly an inch or half an inch, and sometimes a quarter of an inch, subdivided into 100 equal parts. To construct this scale, draw II equidistant parallel lines ; divide the upper of these lines A E into such a number of equal parts as the scale is intended to contain; from each of these divisions draw perpendicular lines through

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