Elements of geometry and trigonometry . the angles at the bases of the triangles whose vertexis in O, is less than the sum of the angles at the bases of thetriangles whose vertex is in S ; hence to make up the dt-li-cicncy, the sum of the angles formed about the point O, isgreater than the sum of the angles fjrmcd al)out the point the sum of the angles ab(»ut th(. point () is ecjual to foui-rigiit angles (Book Î. Prop. IV. Sch.) : therefore the sum (>f th»plane angles, which Iorm the solid angle S, is less than fourIght angles. Schohum. Tins (i«;iiionslralion i> lounded on the supp
Elements of geometry and trigonometry . the angles at the bases of the triangles whose vertexis in O, is less than the sum of the angles at the bases of thetriangles whose vertex is in S ; hence to make up the dt-li-cicncy, the sum of the angles formed about the point O, isgreater than the sum of the angles fjrmcd al)out the point the sum of the angles ab(»ut th(. point () is ecjual to foui-rigiit angles (Book Î. Prop. IV. Sch.) : therefore the sum (>f th»plane angles, which Iorm the solid angle S, is less than fourIght angles. Schohum. Tins (i«;iiionslralion i> lounded on the suppositionthat the solid angle is convex, or that the plane of no one sur-face prfMluced can ever nnct the s(jlid angle ; if it were other-wise, the sum of the jdanc angles would no longer be limit(»<l,and might be of any magnitude. XXI. THEOREM. Iflwo solid angles arc cuiitnincd Inj three plan», angles w/iuli arcequal to each other, each to cach^ the planes of the equal angleswill be equally inclined to each HÙ GEOMETRY. Let the angle ASC=DTF,theangle ASB=:DTE, and the an-gleBSC=ETF; then will theinclination of the planes ASC,ASB, be equal to that of theplanes DTP, DTK Having taken SB at pleasure,draw BO perpendicular to theplane ASC ; irom the point O, at which the perpendicularmeets the plane, draw OA, OC perpendicular to SA, SC ;draw AB, BC ; next take TE=SB ; draw EP perpendicular tothe plane DTP ; from the point P draw PD, PF, perpendicularrespectively to TD, TF ; lastly, draw DE, EF. The triangle SAB is right angled at A, and the triangle TDEat D (Prop. VI.) : and since the angle ASB = DT_î: wt^ haveSBA=TED. Likewise SB=TE; therefore the triangle SABis equal to the triangle TDE; therefore SA=TD, and AB=DE,In like manner, it may be shown, that SC=TF, and BC= granted, the quadrilateral SAOC is equal to the quadri-lateral TDPF; for, place the angle ASC upon its equal DTP;because SA=TD, and SC=TF, the point A will fall on D,and the point C
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Keywords: ., bo, bookcentury1800, booksubjectgeometry, booksubjecttrigonometry
