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In the column of versed sines find I
And the corresponding area is

.040875
Sq. diameter,

Area as before, 408.750 000

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Anf. 159.09.

Ex. 2. Required the area of the segment, when the arch is 90° and diameter 36 feet.

Anf. 92.4696 sq. feet. Ex. 3. What is the area of the segment of a circle, when the diameter is 25 and versed sine 9 ?

Ex. 4. Required the area of a segment, whose chord is 32; the radius being 20.

Anf. 178.9168. Ex. 5. Required the area of a segment, its versed fine being 3i, and diameter 50 yards.

Anf. 54.1475 sq. yards.

PROBLEM XXIV.

To find the area of the cycloid.

DeFINITIONS.

1. If the circle ABGE roll on the straight line CD, so that all the points of the circumference be applied to it successively, the point x, that touches the line CD in c; by a motion thus compounded of a circular and rectilineal motion, will describe the curve line CBD, which is called the Cycloid.

2. The straight line CD is called the base.

3. The straight line AB, perpendicular to CD, and bisecting it, is called the axis, and is equal to the diameter of the generating circle.

4. The generating circle is that by whose revolution the curve line is described. 5. The point B is called the vertex.

Note. The base CD is equal to the circumference of the generating circle, and the cycloid CBD is quadruple of the diaVido Sir Isaac Newton's Philosophical Discoveries.

RULE

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meter.

RULE. Multiply the area of the generating circle by 3, and the product is the area of the cycloid.

EXAMPLE I. Required the area of the cycloid, when the diameter of the generating circle is 4 feet.

4 4

16 .7854

47124

7854
12.5664 area of the generating circle.

3
37.6992 sq. feet, area of the cycloid.
Ex. 2. Required the area of the cycloid, whose base is
15.1328.

Anf. 150.79 Ex. 3. Required the area of the cycloid, whose-length is 400 feet.

Anf. 23562 /9. feet.

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PROBLEM XXV.

To find the fine and cofine of a very small arch, such as I', A small arch such as 1', may be considered nearly equal to its fine. Suppose, then, the radius of a circle to be 100000, in which case the circumference will be 628318.52; therefore 628318.52

will quote 29.08, the natural line of i'. Since the бох збо square of the hypothenuse of a right angled triangle is equal to the sum of the squares of the legs, therefore from the square of

the

Y 2

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the radius subtract the square of the fine of any arch, and the. square root of the remainder will be the cofine of that arch.

Thus 10000000000-845.64599999.9 the cosine of 1' The versed fine x B may be found by subtracting the coline, from the radius.

PROBLEM XXVI.

The fine and-coline of any arch being given, to find the fine and co

fine of its double.

RULE.

As the radius is to the cosine of any arch, fo is twice the fing. of that arch to the fine of its double.

EXAMPLE

Required the fine and cosine of two degrees, the fine of

. 1 being 1745, and cosine 99985.

Rad. Co-fine 19
100000:99985:: 3490 : 3489,47 natural fine of 2
3489.472 =

And

If three arches differ equally, the radius is to the cofine of the middle arch as twice the fine of the difference is to the difference of the fines of the greatest and least arches.

Ex. 2. Required the line and cosine of 3°, the fine and com fine of 1° and 2° being given.

100000,: 99939 :: 3490

3490

8994510
399756
299817
3487.87110 =AL the diff of the extreme arches,
1745

Sine of 1°
5232.87110 the sine of zo=AF
The cosine of which is EA-AF2=99863 the cosine of 3o

Ex. 3. The fine and cosine 20 and 3° being given, required the fine and coline of 4°.

Sine 6976
Anf.

Co-sine 99756 Ex. 4. Required the fine and cofine of so, the fine and cofine of 3° and 4° being given.

Sine 8715.4844 Anf

Co-fine 99619 Ex. 5. The fine and cosine of 4° and so being given, requi.. ted the fine and cofine of 6°

Sine

10452 Co-fine

99455

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In like manner, the fine and cosine of every minute and dea gree of the quadrant may be found; but when the calculations are carried on the length of 60°, the fines of the remaining arches may be found by the following rule :

Take the fine of an arch as much below. 60° as the arch whose fine is requiredis above 60°, to which add the fine of the number of degrees that the proposed arch exceeds 60°; the sum will be the fine required. Ex. 6. What is the line of 80°?

The fine of noo-20o=40° is 64279
The line of 80°-60°=20° is 34202

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Answers.
Ex 7. Required the fine of 70° 15'

94118
8.

of 67° 14

92209
9.

of 79° 30' 98325
of 89° 45'

99999 The versed sine is found by subtracting the cosine from the ra. dius.

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PROBLEM XXVII.

| To find the tangent and cotangent of every minute and degree of the

quadrant, the fines and cosines being given. The tangent and cotangent of any arch may be found by either of the following proportions : Because the triangles CED and CBA are similar, CD : DE::.

DExCA ÇA : AB, therefore --- = AB, the tangent of the arch,

CD EA ; that is, the rectangle contained by the fine and radius of any arch, is equal to the rectangle contained by the coline and tangent of that arch. Hence,

RULE I.

To find the tangent, multiply the fine of any arch by the radius, and divide the product by the coline, the quotient will be the tangent of that arch. The cotangent

of
any
arch
may
be fouad

upon

the same prin

LEX CF ciples : Thus, CL: LE :: CF : FK ; therefore,

=FK;

CL hence,

RULE 2. Divide the product of the cosine and radius of anys arch by its fine, and the quotient will be the cotangent: or (which is the same thing) fay, As the line of any arch is to its co-line: so is the radius to the cotangent of that arch.

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