6. A surveyor, wishing to determine the side AB of a field, rendered incapable of direct measurement by reason of an intervening morass, runs the line AC, south 38° west, 7.75 chains, then CB, south 25'8 east, 10'15 chains.

=

Produce AC in K and draw the meridian CS; then SCK = 38° and BCS = 25'8°,

(CAB+CBA) = BCK = 31.9°.

K

The angle A being determined the solution will be readily finished, and we shall find

AB, S 1° 19'′ 14′′ W, 15'246 chs.

[Verify by employing the computed value of DA to find AB already known.]

CASE IV. The sides given, to determine the angles.

Rule 1. log. sin+A=[log. (h- b)+log. (h-c)-(log. blog. c)]. 2. log. cosA [log. h+log. (h-a)-(log. b+log. c)]. 3. log. tan A=log. (h-b)+log. (h-c)-log. h-log. (h-a)]. 4. log. sinA [log. h+log. (h-a)+log. (h-b)+log. (h-c)] +log. 2-(log. b+log. c).

8°. Let it be required to find the angles of a triangular field, the sides of which are 3004, 2674, 199 feet.

The third operation for the computation of C is obviously un

necessary, unless we wish to test the accuracy of the work. We have employed three methods in order to illustrate the rules; sometimes one and sometimes another will be preferable, according to the numbers.

9°. Required to determine the angles of a quadrilateral field from the following data:

AB=56, BC= 76, CD = 87, DA = 43, BD = 67.

10°. Given two sides of a triangle 367-23, 273 chains, and the difference of the opposite angles 15'7°, to determine the triangle. 11°. Given the sum of two arcs and the ratio of their sines, to determine the arcs.

Let the arcs be denoted by u, v, their sum by e, and their ratio by r; we have

whence u and v are finally determined.

The student may make an application of the above in the solution of the following problem, taken from Davies' Legendre:

12°. "From a station, P, there can be seen three objects, A, B, and C, whose distances from each other are known, viz., AB=800, AC = 600, and BC= 400 yards. There are also measured the horizontal angles, APC = 33° 45', BPC = 22° 30′. It is required from these data to determine the three distances PA, PC, and PB.”

The angles CAP and CBP will be the u and v of the eleventh. The student will make the computations, and devise means to satisfy himself of the correctness of his results.

13°. Wishing to ascertain the distance, AP, to an inaccessible object, P, also invisible from A, I measure to the right and left the

=

equal lines AB AC = 2137 chains, and the angles, BAC 113° 12', ABP = 65° 36', ACP = 89° 5'.

=

How might the principle of this problem be applied to determine the distance of the moon, her zenith distances being observed by two astronomers, one at St. Petersburg, and the other at the Cape of Good Hope?

SECTION THIRD.

Quadrature of the Circle, the Ellipse, and Parabola.

PROPOSITION I.

To find the area of a circular sector in terms of its radius and arc.

The sector y is obviously a function of its arc x, the radius, r, being a constant quantity. Giving to x and y the vanishing increments, h and k, we have (311),

.. (246),

but

[k] = {r • [h];

y' = [ '1⁄2 ] = + r = +rx°;

y=rx + constant,

Yz=0 = 0;

y = rx, i. e.,

Fig. 64.

The circular sector is measured by half the product of (397) its radius and arc.

Cor. The area of a circle is equal to the product of the (398) radius multiplied into its semicircumference.

Scholium. The celebrated Problem of the Quadrature of the Circle is evidently reduced to the following proposition:

PROPOSITION II.

The diameter of a circle being given, it is required to find the circumference.

* See Montucla, "Histoire des Recherches sur la Quadrature du Cercle."

So

If in (371) we make xr, there results

y = circumference = (1 − + − ‡ +, −, ...)r ;
(circumference), = 4(1 − ↓↓↓ — 4+, −, ...) • 2r.
(circumference), 4(1-+-++, −, ...) • 21°2,

for any other radius, r2;

(circumference),

2

(circumference)TM1⁄2

=

and if 2r,= 2, and .. (373), (circ.), = 2π, we have

The circumference of any circle bears to its diameter (399) a constant ratio.

(circumference), =π 2r=3.1415926535897993. 2r.

Cor. 1. The arcs of similar sectors are to each other as (400) the radii of the respective circles of which they form like parts. For, if u, u, denote the arcs of similar sectors, or like parts, as the nth, of the circumferences of which the radii are r, r2, we have

Eliminating the semicircumference from (398) by aid of (399), there results,

Cor. 2. Area of circle = r2; or, the circle is measured (401) by the square of the radius multiplied into T.

Cor. 3. Circles and the like parts of circles, as similar (402) sectors and segments, are to each other as the squares of their proportional lines-such are the radii, diameters, circumferences, similar arcs and their chords, sines, tangents, secants, and versedsines.

For let Ss and S2+ s2 be similar sectors with their equal angles made vertical, s, s, the similar segments cut off from the sectors by the chords c, c r, ra, being the radii. Then S+s, S2+ S2, being like parts of their respective circles (?) if n(S+s) represent the first, n(S,+ s2) will represent the second circle; and we shall have (401)

S

Fig. 66.