. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. sin. q-\-$hy sin. p=^xy sin. A r\ i aw /0x Or # sin. <H-y sin. p— - (2) h If we had required the rath part of the triangle AB C, in place of the 3rd part, equation (1) would have been nxy=bc. Making this supposition, to make the problem more general, we have xy= - and y=. — By the aid of these last two equations,ra nx (2) becomes , be sin. p be sin. A x sin. q-f- - = nx np ~ be sin. A be sin. p Or, a;*—__; x=— :-£ ,3x raA sin. q rasin. ^ 1°; Whence, ?-bc s
. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. sin. q-\-$hy sin. p=^xy sin. A r\ i aw /0x Or # sin. <H-y sin. p— - (2) h If we had required the rath part of the triangle AB C, in place of the 3rd part, equation (1) would have been nxy=bc. Making this supposition, to make the problem more general, we have xy= - and y=. — By the aid of these last two equations,ra nx (2) becomes , be sin. p be sin. A x sin. q-f- - = nx np ~ be sin. A be sin. p Or, a;*—__; x=— :-£ ,3x raA sin. q rasin. ^ 1°; Whence, ?-bc sin ^=b/ 52c2sin-2^ _^ \4 n2h2 q rasin. ^/ In case ra is Zar^e, that is the part to be cut off small, the valueof x may be imaginary, corresponding to the preceding remark. Case 5. When the given point is on one side of the triangle. The two parts must be equal, or oneof them will be less than half of thewhole. We always compute the less part. LetP be the point in the line AB, P Q andPR perpendiculars to the other sides, areknown ; B C and A G are both , through the given point P, it is. 128 SURVEYING. required to draw PD, so that the triangle BPJ), shall be the nthpart of ABC. That is BCAG PQ. BZ>=: Whence, BD-. nPQ Case. 6. When the given point is without the triangle. Let ABC be the given triangle, and P any given point withoutit. It is required to run a line from P, to cut off a given portion ofthe triangle ABC, or (which is the same thing) to divide the tri-angle into two parts having the ratio of m to n. Let PG be theline required. As P is a given point, AP is a linegiven in distance and position; therefore,the angle PAHis known. Solution. — Put the angle PAH—u,CAB=v; then PAG=(u+v). Alsoput AG=x, AR=y, AP—a, and thearea of the triangle AHG=mc, mc being a known quantity. Now, (by Prob. III. Mens.) \xy sin. v=mc (1) Also \ax sin. (w-}-v)=area APG And \ay sin. w=area PAH Therefore, \ax sin. (u-{-v)—\ay sin. u=mc (2)
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Keywords: ., boo, bookcentury1800, booksubjectnavigation, booksubjectsurveying
