cos (A+B): cos(AB) :: tang c: tang (a+b), sin (A+B) sin (A - B) B): tangc: tang (a - b).

From which a and b are found as in the last case. The remaining angle can then be found by Case I.

A

=

Ex. 1. In a spherical triangle ABC, there are given 81° 38′ 20′′, B = 70° 09′ 38′′, c = 59° 16′ 23′′: to find the remaining parts.

}(A+B)=75° 53′ 59′′, }(A—B)=5° 44′ 21′′, }c=29° 38′ 11'.

COS 1⁄2 (A+B) 75° 53′ 59′′ log. ar. comp. 0.613287

: COS

(AB) 5° 44′ 21′′

9.997818

Ex. 2. In a spherical triangle ABC, there are given

= B= A 34° 15′ 03′′, B = 42° 15 13", and c = 76° 35′ 36′′:

MENSURATION OF SURFACES.

1. WE determine the area, or contents of a surface, by finding how many times the given surface contains some other surface which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should understand that one square foot is taken for the unit of measure, and that this unit is contained 9 times in the square yard.

2. The most convenient unit of measure for a surface, Is a square whose side is the linear unit in which the linear dimensions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &c.

3. We have already seen (B. IV., P. 4, s. 2), that the term, rectangle or product of two lines, designates the rectangle constructed on the lines as sides; and that the numerical value of this product expresses the number of times which the rectangle contains its unit of measure.

4. To find the area of a square, a rectangle, or a parallel

ogram.

Multiply the base by the altitude, and the product will be the area (B. iv., P. 5).

Ans. 104.125.

Ex. 1. To find the area of a parallelogram, the base being 12.25, and the altitude 8.5. 2. What is the area of a square whose side is 204.3 feet? Ans. 41738.49 sq. ft. 3. What are the contents, in square yards, of a rectan gle whose base is 66.3 feet, and altitude 33.3 feet?

Ans. 245.31.

4. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 sq. ft. 5. To find the number of square yards of painting in a parallelogram, whose base is 37 feet, and altitude 5 feet 3 inches. Ans. 21.

5. To find the area of a triangle.

CASE I.

When the base and altitude are given.

Multiply the base by the altitude, and take half the product. Or, multiply one of these dimensions by half the other (B. IV., P. 6).

Ex. 1. To find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. To find the number of square yards in a triangle, whose base is 40, and altitude 30 feet. Ans. 663.

3. To find the number of square yards in a triangle, whose base is 49, and altitude 251 feet. Ans. 68.7361.

CASE II.

6. When two sides and their included angle are given. Add together the logarithms of the two sides and the logarith mic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area. Let BAC be a triangle, in which

there are given BA, BC, and the included angle B.

From the vertex A draw AD perpendicular to the base BC, and repre

B

D

sent the area of the triangle by Q. Then (Trig. Th. I.), sin B :: BA : AD;

R:

hence, by substituting for AD its value, we have,

Taking the logarithms of both members, we have, log. 2 Q = log. BC + log. BA + log. sin Blog R; the formula of the rule as enunciated.

Ex. 1. What is the area of a triangle whose sides are, BC = 125.81, BA = 57.65, and the included angle B 57° 25'?

=

and 2 Q = 6111.4, or Q3055.7, the required area.

2. What is the area of a triangle whose sides are 30 and 40, and their included angle 28° 57'?

Ans. 290.427.

3. What is the number of square yards in a triangle of which the sides are 25 feet and 21.25 feet, and their included angle 45°? Ans. 20.8694.

CASE III.

7. When the three sides are known.

1. Add the three sides together, and take half their sum. 2. From this half-sum subtract each side separately. 3 Multiply together the half-sum and each of the three re mainders, and the product will be the square of the area of the triangle. Then, extract the square root of this product, for the required area.

Or, After having obtained the three remainders, add together the logarithm of the half-sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area.

Let ACB be a triangle: and denote

the area by Q: then, by the last case, we have,

By substituting in this equation the values of sin†A, and cos A, found in Arts. 92 and 93, Plane Trigonometry, we obtain,

Ex. 1. To find the area of a triangle whose three sides 20, 30, and 40.

The square root of which is 290.4737, the required area.

2. How many square yards of plastering are there in a triangle whose sides are 30, 40, and 50 feet?

8. To find the area of a trapezoid.

Ans. 663.

Add together the two parallel sides: then multiply their sum by the altitude of the trapezoid, and half the product will be the required area (B. IV., P. 7).

Ex. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 152075.

2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches?

Ans. 1313 sq. ft.

3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet?

Ars 20531.