. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. -v)=area APG And \ay sin. w=area PAH Therefore, \ax sin. (u-{-v)—\ay sin. u=mc (2). Or, / . \ • 2 mc?sin. \u-\~v)—ysm. %= a (3) From (1), we find y- mc a: sin. v which value substituted in (3) gives . / , x 2 mc sin. u x sin. (u-\-v)— . : x sm. v c. Whence, x2 sin. (w-j-v)— - 0 jzmca mc 2 mc sin. u—x= . a sin. v 2 mc sin. w r, a sin. (u-\-v) sin. v sin. (u-\-v) mc r Therefore,*-, sin. (M+t>)±V - m2c2 +< 2 mc sm. w (u-\-v) sin. v sin. («-J-v) DIVISION OF
. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. -v)=area APG And \ay sin. w=area PAH Therefore, \ax sin. (u-{-v)—\ay sin. u=mc (2). Or, / . \ • 2 mc?sin. \u-\~v)—ysm. %= a (3) From (1), we find y- mc a: sin. v which value substituted in (3) gives . / , x 2 mc sin. u x sin. (u-\-v)— . : x sm. v c. Whence, x2 sin. (w-j-v)— - 0 jzmca mc 2 mc sin. u—x= . a sin. v 2 mc sin. w r, a sin. (u-\-v) sin. v sin. (u-\-v) mc r Therefore,*-, sin. (M+t>)±V - m2c2 +< 2 mc sm. w (u-\-v) sin. v sin. («-J-v) DIVISION OF LANDS. 129 EXAMPLES. 1. In the triangle ABC, the side . chains, AC=\chains, and BC= chains. The given point P from the angle A, is distant 10 chains, at anangle of 40° from the line AC. It is required to draw a line from this given point P, through thetriangle, so as to divide it into two equal parts. Whereabouts on AB will P G intersect ? The angle BAC=31° 17 19=v. PAH=40°=u. ThereforePAG=1\° 17 10=(w-K) The area of the triangle ABC is square chains. Thepart to be cut off by the triangle AHG is therefore = We must use the natural sines, or the logarithmic
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Keywords: ., boo, bookcentury1800, booksubjectnavigation, booksubjectsurveying
