. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. by sine gives cotangent. Thus we shall 0btain J< -L 1 \=3d (5) \cot. P—cot. v/ \cot. P—cot. uy This last equation shows the surveyor that if he can make it conven-ient to run his random line from P, perpendicular to one of the sides,his equation will be less complex. For instance, if A IIP=90°, itscotangent will be 0, and cot. u would then =0, and equation (5)would become. b2 a2 cot. P—cot. tT^cot. PFor the sake of convenience put cot. P=z, and cot. v=
. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. by sine gives cotangent. Thus we shall 0btain J< -L 1 \=3d (5) \cot. P—cot. v/ \cot. P—cot. uy This last equation shows the surveyor that if he can make it conven-ient to run his random line from P, perpendicular to one of the sides,his equation will be less complex. For instance, if A IIP=90°, itscotangent will be 0, and cot. u would then =0, and equation (5)would become. b2 a2 cot. P—cot. tT^cot. PFor the sake of convenience put cot. P=z, and cot. v= ?.2d (6) Or, b2 a^ z—c z -b* W4)r a2c 2d V %d The numerical value of z will be the numerical value of cot. P; itslogarithm taken, and 10 added to the index will be logarithmic our table. The same remarks will apply to cot. v or c. Case 3. When the given point is without thepolygon. Let ABCDE be the polygon, and P thegiven point without it. From the last case we learn that the surveyorhad better run his random line perpendicularto one of the sides, therefore let PUG be therandom line, perpendicular to DIVISION OF LANDS. 139 As before, compute the area, AEGH, subtract it from mc, the dif-ference is the difference between the triangles PGL and PKL the line that divides the polygon as required. Put PH=a, PG=b, PK=x, PL=y, angle ZT=90°, angle PGEs=w, and the angle at P, designated by P. The triangle PHK=\ax sin. PPGL=\by sin. PWhence, b sin. Py—a sin. P x=2d. (1) Here (2) represents a similiar quantity as in the last case. In the triangle PHK, we have 1 : x : : cos. P a, or x= COS. P In PGL, sin. u : y : : sin. (w—P) : 5. b sin. w (2)(3) sin.(w—P) When the values of x and y are substituted in (1) we havesin. P sin, u sin. Psin. (w—P)~a2c^s7P:= (4) / sin. P sin. u \ sin. P \sin. u cos. P—cos. u sin. P/ a cos. P b2 a2 0r> cot. P—~~cot7P=2^ (6) This equation is exactly similar to equation (6) of the last case, M 2d (5) and it is redu
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Keywords: ., boo, bookcentury1800, booksubjectnavigation, booksubjectsurveying
