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Divide the versed sine by the diameter, find the quotient in the column of versed fines, and multiply the corresponding area by the square of the diameter for the area of the segment.

The example being the fame as before, we have the versed fine equal 10, and diameter 100.

100)10.0(.1

100

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Ex. 2. Required the area of the segment, when the arch is

90° and diameter 36 feet.

Anf. 92.4696 fq. feet.

Ex. 3. What is the area of the segment of a circle, when the Anf. 159.09.

diameter is 25 and versed sine 9?

Ex. 4. Required the area of a segment, whose chord is 32, Anf. 178.9168.

the radius being 20.

Ex. 5. Required the area of a segment, its versed sine being Anf. 54.1475 Sq. yards.

34, and diameter 50 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 č, 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 defcribed.

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 diameter. Vido Sir Ifaac Newton's Philofophical Discoveries.

RULE RULE.

Multiply the area of the generating circle by 3, and the pro

duct is the area of the cycloid.

EXAMPLE 1.

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 fq. feet, area of the cycloid.

Ex. 2. Required the area of the cycloid, whose base is 25.1328. Anf. 150.79. Ex. 3. Required the area of the cycloid, whose length is 400 feet. Ans. 23562 sq. feet.

PROBLEM XXV.

To find the fine and cofine of a very small arch, fuch as I'

A small arch such as 1', may be confidered nearly equal to its fine. Suppose, then, the radius of a circle to be 100000, in which cafe the circumference will be 628318.52; therefore. 628318.52

will quote 29.08, the natural fine of 1'. Since the

60 × 360 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 radius fubtract the square of the fine of any arch, and the square root of the remainder will be the cofine of that arch.

Thus 10000000000-845.64=99999.9 the cofine of I' The versed sine x B may be found by fubtracting the cosine, from the radius.

PROBLEM XXVI.

t

The fine and cofine of any arch being given, to find the fine and cofine of its double.

RULE.

As the radius is to the cosine of any arch, so is twice the fine

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 1o

100000:99985:: 3490: 3489.47 natural fine of 2

3489.472 =

AndV

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 fine and cosine of 3o, the fine and cofine of 1o and 2o 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 fine of 3°=AF

The cosine of which is EA2-AF2=99863 the cofine of 3o

Ex. 3. The fine and cosine 2o and 3o being given, required the fine and cofine of 4°.

1

Anf.

Sine 6976
Co-fine 99756

Ex. 4. Required the fine and cosine of 5°, the fine and co

fine of 3o and 4o being given.

Sine 8715-4844
Anfin
Co-fine 99619

Ex. 5. The fine and cosine of 4o and 5o being given, requi red the fine and cosine of 6°

Ans.

Sine 10452 Co-fine 99455

In like manner, the fine and cosine of every minute and degree 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 sine is required is above 60°, to which add the sine of the number of degrees that the proposed arch exceeds 60°; the sum will be the fine required.

Ex. 6. What is the fine of 80°?

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

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