. Elements of plane and spherical trigonometry . by hypothesis polars. tHence B is the pole of ca, thatis of ac; A is the pole of be,that is of be; and since C isthe pole of ab, therefore C isthe other pole of ab. Thepoints A, B, C^ are the poles ofthe sides, therefore, of the tri-angle abc. Whence, also, itfollows by the reciprocity ofthe polar system, that a, b, care the poles of ABC. In the same way it is shownthat ACB and acb form apolar system, and that BA^C,hac, form another. Hence it follows that if oneciated system be polar to one of another associated system, then eachof the triangles
. Elements of plane and spherical trigonometry . by hypothesis polars. tHence B is the pole of ca, thatis of ac; A is the pole of be,that is of be; and since C isthe pole of ab, therefore C isthe other pole of ab. Thepoints A, B, C^ are the poles ofthe sides, therefore, of the tri-angle abc. Whence, also, itfollows by the reciprocity ofthe polar system, that a, b, care the poles of ABC. In the same way it is shownthat ACB and acb form apolar system, and that BA^C,hac, form another. Hence it follows that if oneciated system be polar to one of another associated system, then eachof the triangles of one system is polar to one triangle of the other asso-ciated system. * In the Edinburgh Transactions the author of this supplement has investigatedLexells theorem by a novel analysis,—fAe geometry of spherical coordinates, whichdetermines in a direct manner the equation of the locus, and then shows its identitywith the previously found equation of a circle. t When we speak of parallel lines without specifying which is taken as the line of. triangle of an asso- i^ SPHERICAL GEOMETRY. This is not the only cnrious property of the figure before us: and weshall put down a small selection from those we are in possession of,not doubting that on many accounts they will be interesting to thosegeometers who indulge in trigonometrical speculations. An addition tothese will appear in the 25th number of the Mathematical Repository. Draw the great circle aA. (next figure), and produce it to met BC,be in G and H. Then, because a is the pole of BC, and that A is thepole of be, the arcs aG, AH, are quadrants. Hence, aQ-\- AH = \- AG = IT, and the angles at G and H are right angles. / Let R, R, be the intersections of BC, he ; then, because xUe angles atG and H are right, R, R, are the poles of a A. Let AB meet be in K, BC meet ab inL, AC meet be in M, and ae meet Be in N. ThenZ>H-^Hc = GAC^GAB ) &H-I-GAB He + GAC. For JM = Z»H+ HM = 6H + HAM= ^H -f- GAC. In like manner, cK —cH +
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