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The length of the line 75 feet, being divided by 25, will give 3, the number of inches which will represent the line on paper

Therefore, draw the indefinite line AB, on which lay off a

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distance AC equal to 3 inches: AC will represent the given line of 75 feet drawn to the required scale.

REMARK I. This problem explains the manner of laying down a line upon paper, in such a manner that a given number of parts shall correspond to the unit of the scale, whether that unit be an inch or any part of an inch.

When the length of the line to be laid down is given, and it has been determined how many parts of it are to be represented on the paper by a distance equal to the unit of the scale, we find the length which is to be taken from the scale by the following

RULE.

Divide the length of the line by the number of parts which is to be represented by the unit of the scale : the quotient will show the number of parts which is to be taken from the scale.

EXAMPLES.

1. If a line of 640 feet in length is to be laid down on paper, on a scale of 40 feet to the inch; what length must be taken from the scale ?

40)640(16 inches. 2. If a line of 357 feet is to be laid down on a scale of 68 feet to the unit of the scale, (which we will suppose half an inch), how many parts are to be taken?

Ans. (6.85; parts, or REMARK II. When the length of a line is given on the paper, and it is required to find the true length of the line which it represents, take the line in the dividers and apply it to the scale, and note the number of units, and parts of an

unit to which it is equal. Then multiply this number by the number of parts which the unit of the scale represents, and the product will be the length of the line.

For example, suppose the length of a line drawn on the paper was found to be 3.56 inches, the scale being 40 feet to the inch: then,

3.55 X 40=142 feet, the length of the line.

PROBLEM VIII.

Having given two sides and the included angle of a triangle, ro

describe the triangle.

B
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150

120

34. Let the line B=150 feet, and C=120 feet, be the given sides; and

A A=30 degrees, the given angle: to

H describe the triangle on a scale of 200 feet to the inch.

Draw the indefinite line DG, and at the point D, make the angle GDH equal to 30 degrees ; then lay off DG equal to 150, equal to three quarters of an inch, and DH equal to 120, equal to six tenths of an inch, and draw GH: DGH will be the required triangle.

PROBLEM IX.

The three sides of a triangle being given, to describe the

triangle.

35. Let A, B and C, be the sides. Draw DE equal to the side A. From the point D as a centre, with a radius

Dequal to the second side B, describe an

AH arc : from E as a centre, with a radius equal to the third side C, describe CH another arc intersecting the former in F; draw DF and EF, and DEF will be the triangle

BH

PROBLEM X.

Having given two sides of a triangle and an angle opposite one

of them, to describe the triangle.

36. Let A and B be the given

AH sides, and C the given angle which

a

BH we will suppose is opposite the side B. Draw the indefinite line DF and

H make the angle FDH equal to the

D

F angle C: take DH=A, from the point H, as a centre, with a radius equal to the other given side B, describe an arc cutting DF in F; draw HF: then will DHF be the required triangle.

If the angle C is acute, and A

C the side B less than A, then the BH arc described from the centre E with the radius EF = B will cut the side DF in two points, F and D G, lying on the same side of D: hence there will be two triangles, DEF, and DEG, either of which will satisfy all the conditions of the problem.

PROBLEM XI.

The adjacent sides of a parallelogram, with the angle which

they contain, being given, to describe the parallelogram.

37. Let A and B be the given sides, F and C the given angle.

Draw the line DE=A; at the point D D, make the angle EDF=C; take AH

lo DF=B: describe two arcs, the one

BH from F, as a centre, with a radius FG=DE, the other from E, as a centre, with a radius EG=DF; through the point G, where these arcs intersect each other, draw FG, EG; DEGF will be the parallelogram required.

PROBLEM XII.

To find the centre of a given circle or arc.

38. Take three points, A, B, C, any where in the circumference, or in the arc: draw AB, BC; bisect these two lines by the perpendiculars, DE, FG: the point where these perpendiculars meet will be the centre sought.

The same construction serves for making a circumference pass through three given points A, B, C, and also for describing a circumference, about a given triangle.

CHAPTER III.

Plane Trigonometry.

39. In every plane triangle there are six parts: three sides and three angles. These parts are so related to each other, that if a certain number of them are known or given, the remaining ones can be determined.

40. Plane Trigonometry explains the methods of finding, by calculation, the unknown parts of a triangle when a sufficient number of the six parts is given.

It has already been shown, in the problems, that triangles may be constructed when three parts are known. But these constructions, which are called graphic methods, though perfectly correct in theory, would give only a moderate approxi. mation in practice, on account of the imperfection of the in struments required in constructing them.

Trigonometrical methods, on the contrary, being inde. pendent of mechanical operations, give solutions with the utmost accuracy.

41. For the purposes of trigonometrical calculations, the circumference of the circle is divided into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes ;

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As the circumference of a circle may be regarded as a proper measure of angles, having their vertices at the centre, the four right angles which can be formed about the same point, are measured by 360 degrees; two right angles by 180 de grees, one right angle by 90 degrees, and an angle less than a right angle, by an arc less than 90 degrees.

Degrees, minutes, and seconds, are usually designated by the respective characters, '". Thus, 16° 12' 15" is read, 16 degrees, 12 minutes, and 15 seconds. 42. The complement of an arc is L

N what remains after subtracting the arc from 90°. Thus, the arc EB is the complement of AB. The sum of an arc and its complement is equal o

H

D to 90'.

43. The supplement of an arc is what remains after subtracting the arc from 180°. Thus, GF is the sup

Q plement of the arc AEF. The sum of an arc and its supplement is equal to 180°.

44. The sine of an arc is the perpendicular lét fall from one extremity of the arc on the diameter which passes through the other extremity. Thus, BD is the sine of the arc AB.

45. The cosine of an arc is the part of the diameter intercepted between the foot of the sine and centre. Thus, OD is the cosine of the arc AB.

46. The tangent of an arc is the line which touches it at one extremity, and is limited by a line drawn through the other extremity and the centre of the circle. Thus, AC is the tangent of the arc AB.

47. The secant of an arc is the line drawn from the centre of the circle through one extremity of the arc, and limited by the tangent passing through the other extremity. Thus, OC is the secant of the arc AB.

48. The four lines, BD, OD, AC, OC, depend for their values on the arc AB and the radius OA; they are thus designated :

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