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known to be 3.14159. This being divided successively by 180 and 60, gives .0002908882 for the arc of one minute, which may be regarded as the sine of one minute.

The cosine of 1=V1-sin."=0.9999999577. The sines of very small angles are nearly proportional to the angles themselves. We might then obtain several other sines by direct proportion. This method will give the sines correct to five decimal places, as far as two degrees. By the following method they may be obtained with greater accuracy for the entire quadrant. By Art. 75, we have, by transposition,

sin. (a+b)=2 sin. a cos. b-sin. (ab),

cos. (a+b)=2 cos. a cos. b-cos. (ab). If we make a=b, 26, 3b, &c., successively, we shall have

sin. 2b=2 sin. b cos. b;
sin. 36=2 sin. 2b cos. b-sin. b.
sin. 4b=2 sin. 3b cos. b-sin. 26,
&c.

&o.
cos. 2b=2 cos, b cos. b-1;
cos. 3b=2 cos. 2b cos. b-cos. b;
cos. 4b=2 cos. 3b cos. b-cos. 25,

&o
Whence, making b=1', we have
sin. 2=2 sin, l' cos. 1'

=.000582;
sin. 3=2 sin. 2' cos. 1'-sin. 1'=.000873;
sin, 4'=2 sin. 3' cos. 1'-sin. 2'=.001164,
&c.,

&c.
cos. 2=2 cos. 1' cos. 1'- 1 =0.999999;
cos. 3'=2 cos. 2' cos. 1'--cos. 1'=0.999999;
cos. 4'=2 cos. 3' cos. 1'-cos. 2'=0.999999,
&c.,

&c. The tangents, cotangents, secants, and cosecants are easily derived from the sines and zosines. Thus, sin. 1'

cos. 1!
i

cot. 1'=
cos. 1
1

1
sec. 1'=

Lii
cosec. 1'-

sin. 1'
&c.

&c.

&c.,

1

tang. 1'=

sin. 1';

cos. 1°

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BOOK III.

MENSURATION OF SURFACES.

(80.) The area of a figure is the space contained within the line or lines by which it is bounded. This area is determined by finding how many times the figure contains some other surface, which is assumed as the unit of measure. This unit is commonly a square; such as a square inch, a square foot, a square rod, &c.

The superficial unit has generally the same name as the linear unit, which forms the side of the square. Thus, the side of a square inch is a linear inch l;

of a square foot is a linear foot;

of a square yard is a linear yard, &c. There are some superficial units which have no correspond. ing linear units of the same name, as, for example, an acre.

The following table contains the square measures in com

mon use :

Table of Square Measures.

Sq. Inches.

Sq. Feet. 144=

1

Sq. Yards. 1296=

9

1

Sq. Rods. 39204= 27215 301=

1

S. Ch's. 627264= 4356

484

16= 1 6272640= 43560 4840 160= 10= 1

M. 4014489600=27878400 =3097600 =102400=6400=640=1

Acres.

PROBLEM I.

(81.) To find the area of a parallelogram.

RULE I. Multiply the base by the altitude.

For the demonstration of this rule, see Geometry, Prop. 5, B. IV.

.

Ex. 1. What is the area of a parallelogram whose base is 17.5 rods, and the altitude 13 rods?

Ans., 227.5 square rods. Ex, 2. What is the area of a square whose side is 315.7 feet?

Ans., 99666.49 square feet. Ex. 3. What is the area of a rectangular board whose length is 15.25 feet, and breadth 15 inches?

Ans., 19.0625 square feet. Ex. 4. How many square yards are there in the four sides of a room which is 18 feet long, 15 feet broad, and 9 feet high?

Ans., 66 square yards. (82.) If the sides and angles of a parallelogram are given, the perpendicular height may be found by Trigonometry. For DE is one side of a right-angled triangle, of which AD is the hypothenuse. Hence, R: AD :: sin. A :DE;

ADX sin. A from which

DE=

R

ABX ADX sin. A Therefore, the area=ABX DE

R Hence we derive

D

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А

E

B

RULE II. Multiply together two adjacent sides, and the sine of the included angle.

Ex. 1. What is the area of a parallelogram having an angle of 58°, and the including sides 36 and 25.5 feet? Ans. The area=36X25.5.84805 (natural sine of 58°)=

778.508 square feet. The computation will generally be most conveniently performed by logarithms.

Ex. 2. What is the area of a rhombus, each of whose sides is 21 feet 3 inches, and each of the acute angles 53° 20'?

Ans., 362.209 feet. Ex. 3. How many acres are contained in a parallelogram one of whose angles is 30°, and the including sides are 25.35 and 10.4 chains ?

Ans., 13 acres, 29.12 rods

PROBLEM II.

(83.) To find the area of a triangle.

RULE I.

Multiply the base by half the altitude.
For demonstration, see Geometry, Prop. 6, B. IV.

Ex. 1. How many square yards are contained in a triangle whose base is 49 feet, and altitude 251 feet?

Ans., 68.736. Ex. 2. What is the area of a triangle whose base is 45 feet, and altitude 27.5 feet?

Ans., 618.75 square feet. . (84.) When two sides and the included angle are given, we may use

RULE II.

Multiply half the product of two sides by the sine of the included angle.

The reason of this rule is obvious, from Art. 82, since a tri. angle is half of a parallelogram, having the same base and al. titude.

Ex. 1. What is the area of a triangle of which two sides are 45 and 32 feet, and the included angle 46° 30' ? Ans. The area=45X16X.725374 (natural sine of 46° 30')=

522.269 feet. Ex. 2. What is the area of a triangle of which two sides are 127 and 96 feet, and the included angle 67° 15' ?

Ans. (85.) When the three sides are known, we may use

RULE III.

From half the sum of the three sides subtract each side sev. erally; multiply together the half sum and the three remain. ders, and extract the square root of the product.

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or

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b'+c-a' ADE

2c But

CD’=AC - AD' ;

(bo+c-a)46*c*(+ca-a?)" henco CD:=b

4c?

4c? V 46°c" (6+c*- a)* CD

2c ABX CD But the area

=1V 46°c" (6+c* — a”)”.

2 The quantity under the radical sign being the difference of two squares, may be resolved into the factors 2bc+(6*+o- a*) and 2bc-(0'+c-a); and these, in the same manner, may be resolved into (b+c+a)x(b+c-a), and

(a+b-c)x(ab+c).

a+b+c Hence, if we put S equal to

we shall have

2 the area=vS(S-a) (S-6) (S--c). Ex. 1. What is the area of a triangle whose sides are 125, 173, and 216 feet? Here S=257,

S-b=84,
S-a=132,

S-C=41.
Hence the area= v 257x132x84X41=10809 square feet.

Ex. 2. How many acres are contained in a triangle whose sides are 49, 50.25, and 25.69 chains ?

Ans., 61 acres, 1 rood, 39.68 perches. Ex. 3. What is the area of a triangle whose sides are 234, 289, and 345 feet?

Ans. (86.) In an equilateral triangle, one of whose sides is

a,

thú expression for the area becomes

Vax ax axja
av3

;

4 that is, the area of an equilateral triangle is equal to on fourth the square of one of its sides multiplied by the square root of 3.

Ex. What is the area of a triangle whose sides are each 37 feet?

Ans., 592.79 feet.

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