Cyclomathesis : or, An easy introduction to the several branches of the mathematics; being principally designed for the instruction of young students, before they enter upon the more abtruse and difficult parts thereof . gle ABC there is given AB, and the angle B, and the ratio of AC to BC, to find the fides. Let fall AD on BC ( produced ); and putAB zzi9 AC = a9 the ratio of AC to CB as I tor, then CB s± ra9 and cof. ACB = c. Then rad.(1) : AC (a) : : (<r) : ca = DC. Then(Geom. II. 22.) bb zz aa + mza + 2*r*tf, whence bb ? b aa — . —. , ana # = rr + 1 + 20* ; vrr + 1 + 2 given two fi
Cyclomathesis : or, An easy introduction to the several branches of the mathematics; being principally designed for the instruction of young students, before they enter upon the more abtruse and difficult parts thereof . gle ABC there is given AB, and the angle B, and the ratio of AC to BC, to find the fides. Let fall AD on BC ( produced ); and putAB zzi9 AC = a9 the ratio of AC to CB as I tor, then CB s± ra9 and cof. ACB = c. Then rad.(1) : AC (a) : : (<r) : ca = DC. Then(Geom. II. 22.) bb zz aa + mza + 2*r*tf, whence bb ? b aa — . —. , ana # = rr + 1 + 20* ; vrr + 1 + 2 given two fides andthe included angle \ to find the area. Let CA, AB and the angle A be given; drawCF perpendicular to AB, and let AB=£, AQzzd9S. ) : sd = CF. Then CF X AB~ — area, or — r: area; that is, half the rectangle 2 ° of the fides multiplied by the fine of the includedangle, gives the area. PROB. CXX. Given all the fides of a trapezium, and two oppq/ite^5- angles 9 to find the area. Let the angles B, D be given, and through theother two angles A, C, draw the diagonal AC. Let •*<2. ^ a o K *&i Y< 3*1+ .bauV Seft. VII. PROBLEMS. 421 Let AB=*f BCzzc, CDzzd, DA=/, S. <B=:^ zz q. Then by the laft problem, the area of 65. the triangle CBA = S-, and the triangle CDA SB ze — *, therefore the trapezium zz ^ v . P R O B. CXXI. In the triangle WNE, there is given the fegment SE, angle WNE, W /£<? ratio of NE / and x ^: «—— n/i + #> + 2/*. 1 +/> Or /£«j,Let / r= Ew^+E> then WN+NE.(px + x) : WN —NE (px*-x) : : / : .£Lzi* = r r pX -\- X P I / zz tang. 4- diff. of the angles W and + 1 Whence the angles W, E ar
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