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the direction of the last side, l, if we observe that, on arriving at its first extremity, our progress will have been farther north than south, or our LATITUDE north, when the sum of the northings shall exceed that of the southings, or when

a+b+...+k' or a cos(a,m)+...+k cos(k, m)> 0, i. e., +; and v. v.; also that our position will be found to the east of the first station, or our DEPARTURE east, when the sum of the eastings shall exceed that of the westings, or when

a"+b" +...+k" or a sin(a,m) +...+k sin(k,m)> 0, and v. v. Scholium II. A part of the tabular labor may be saved by (430) making the projections upon and perpendicular to one of the sides; thus, if we assume a as a false meridian, or make (a,m)=0, (421) becomes

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Scholium III. By eliminating between (420) and (419) any (431) two sides, as k and l, may be determined when all the other quantities are given; but it will be better to operate as if I were a meridian, when (l,m) will = 0, (a,m) = (a,l), (b,m) = (b,l), (c,m) = (c,l), and from (420), (419), there will result

k =

-1=

a sin(a,1)+b sin(b,l)+c sin(c,l) +...+j sin(j,l)

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= acos(a,l) +b cos(b,l) +ċ cos(c,l)+...+k cos(k,l).

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15.75 sin 36'5° - 20 sin 15.3° - 30.25 sin 46°

15'75 cos 36'5° +20 cos 153° - 30.25 cos 46°

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If, in accordance with the second scholium, we assume a for the meridian, whose bearing is N 365° E, the NW and SE bearings will be increased by 36'5°, and the NE and SW diminished by the same angle; hence Field Notes, No. I., will be transformed into

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The student will calculate d according to this table.

2o. Calculate e of II by (421) and (422).

3o. Calculate ƒ of III twice, by projecting first on b then on c, and balance the errors by adding together and dividing by 2. 4°. Calculate g of IV and verify by a different method.

5°. Find AG in V.

6°. Calculate the diagonal drawn from the first extremity of a in IV to the second extremity of c.

7°. Solve VI, finding also the length, 7, of the intercepted meridian. [l intersects e: why ?]

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AREAS.

-115 sin69°-17 sin2844°+11 sin31'6°+15 sin56.8°

- sin67.1°

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7 = 11'5 cos69° +17 cos28-4° +11 cos31'6° + 15 cos56'8°

- 0.55 cos67'1° = 36'444.

8°. Required the length, 7, and second point of intersection with the perimeter, of a line running from (a,b) of IV, N 25° W.

PROPOSITION II.

It is required to find the area of a polygon in terms of the sides and the angles which these sides make with each other.

The easiest way of ascertaining the area of any polygon (a, b, c, ..., j, k, l), is obviously to divide it into triangles (A, b), (A, c), (A, d), ..., (A, j), (A, k), and then to compute these triangles. But the double area of a triangle is had at once by taking the product of its base and altitude. Therefore from A, the common vertex of all the triangles, and upon their bases, b, c, d, ..., j, k, pro

k

Fig. 102.

duced, if necessary, let fall the corresponding perpendiculars, Po Pc Par ..., P Pk; there results

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But p, is obviously the projection of a upon the line of P, F. the sum of the projections of a and b upon the line of p., p the sum of the projections of a, b, c, upon the line of Pa ....., P, the sum of the projections of a, b, c, ... i, upon the line of p1, P the sum of

the projections of a, b, c, ..., k, upon the line of PK;

Pki

.. p1 = a cos(a,p.) = a sin(a,b), since ≤ (a,p)+(a,b) = 90°; SO Pc = a sin(a,c)+b sin(b,c),

Pa = a sin(a,d)+b sin(b,d)+c sin(c,d),

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P1 = a sin(a,j)+b sin(b,j)+c sin(c,j)+... +i sin(i,j),

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= a sin(a,k)+b sin(b,k)+c sin(c,k) + ... +j sin(j,k);

therefore, substituting above, and adding, there results,

2(A,b) +2(A,c)

2P=+2(A,d) + &c. +2(A,k)

ab sin(a,b)

+ ac sin(a,c) + bc sin(b,c)
=+ad sin(a,d) + bd sin(b,d) + cd sin(c,d)
+ &c., &c., &c.,

(432)

+ak sin(a,k) + bk sin(b,k) + ... +jk sin(j,k);

.. The double area of any polygon is equal to the sum of the products of its sides, save one, multiplied, two and two, into the sines of the angles formed by the sides belonging to the several products.

Cor. 1. The double area of a triangle is equal to the pro- (433) duct of two of its sides multiplied into the sine of the angle included by them. 2tr(a,b,c) = ab sin(a,b).

Cor. 2. The area of a parallelogram is equal to the pro- (434) duct of its dimensions multiplied into the sine of the included angle (433).

Cor. 3. When two triangles have an angle of the one (435) supplementary to an angle of the other, the triangles are to each other as the products of the sides about the supplementary angles (433).

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The area of a triangle is equal to the square root of the continued product of the half sum of the three sides and the three remainders formed by diminishing this half sum by the sides severally. This furnishes a rule very convenient for the application of logarithms.

Cor. 5. Similar polygons and their like segments and sec- (437) tors are to each other as the squares of their homologous lines, whether sides, diagonals, or radii vectores.

For, preserving the notation under (423) and substituting in (432) there results

AREAS.

233

r, sin(a,b)

+r, sin(a,c)+ror, sin(b,c).

2P+ r sin(a,d) +rra sin(b,d) + r.ra sin(c,d) + &c., &c., &c.,

+rsin(a,k)+r ̧rsin(b,k)+...+r ̧r1 sin(j,k) j

Cor. 6. If the polygon is equilateral, that is

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• a2. (438)

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• a2. (439)

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+sin(a,k) + sin(b,k)=...+ sin(j,k) (

Cor. 7. If the polygon be regular, that is, have its angles as well as its sides equal, putting****

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(c,k) = (n − 3)e,

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(j,k) = e,

n+1 denoting the number of sides of the polygon, there results,

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+sin(n-1)e+sin(n-2)e+sin(n-3)e+...+sine

or 2P+1= [(n - 1)sine + (n-2)sin2e

+(n-3)sin3e+...+ sin(n - 1)e] a2.
+sin(n-1)e]•

Cor. 8. Making n = 2, we find

P3

} (440)

+[(2-1)sin 360°] • a2 = ‡sin60° • a2 = 3*(‡a)2, (441)

for the area of the Equilateral Triangle.

Cor. 9. Making n = 3, we find

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P[2sin90° + sin 180°] a = a2,

for the area of the Square, as already known.

(442)

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