<prompt> Katika mchoro ulio hapa chini, AB AC 115, AD 38, na CF 77. Kokotoa fracCEFDBE.asyukubwa wa kitengo sentimita 0.025;jozi A, B, C, D, E, F;B 0,0;C 80,0;A sehemu ya makutanoarcB,115,0,180,arcC,115,0,180;D interpA,B,38115;F interpA,C,115 77115;E kiendeleziB,C,D,F;kuchoraC--B--A--F--D;leboA, A, N;leboB, B, SW;leboC, C, NE;leboD, D, W;leboE, E, SW;leboF, F, SE;asy  <response>: Kwa kuwa AB 115 na AD 38, BD 115 - 38 77.asyukubwa wa kitengo sentimita 0.025;jozi A, B, C, D, E, F;B 0,0;C 80,0;A sehemu ya makutanoarcB,115,0,180,arcC,115,0,180;D interpA,B,38115;F interpA,C,115 77115;E kiendeleziB,C,D,F;kuchoraC--B--A--F--D;leboA, A, N;leboB, B, SW;leboC, C, NE;leboD, D, W;leboE, E, SW;leboF, F, SE;lebo38, A D2, NW, nyekundu;lebo77, B D2, NW, nyekundu;lebo115, A C2, NE, nyekundu;lebo77, C F2, NE, nyekundu;asyAcha mathbfa iashiria overrightarrowA, n.k. Kwa kuwa AD 38 na BD 77,mathbfd frac77115 mathbfa frac38115 mathbfb.Kwa kuwa AC 115 na CF 77, mathbfc frac77192 mathbfa frac115192 mathbff.Kutenga mathbfa katika kila mlinganyo, tunapatamathbfa frac115 mathbfd - 38 mathbfb77 frac192 mathbfc - 115 mathbff77. Kisha 115 mathbfd - 38 mathbfb 192 mathbfc - 115 mathbff, hivyo 115 mathbfd 115 mathbff 38 mathbf b 192 mathbfc, aufrac115230 mathbfd frac115230 mathbff frac38230 mathbfb frac192 230 mathbfc.Kwa vile viambajengo vya pande zote mbili za mlingano vinaongeza hadi 1, vekta iliyo upande wa kushoto iko kwenye mstari DF, na vekta iliyo upande wa kulia iko kwenye mstari BC. Kwa hivyo, vekta hii ya kawaida ni mathbfe.Zaidi ya hayo, fracEFDE frac115115 1 na fracCEBE frac19238 frac9619 , hivyoanzaalignfracCEFDBE fracfrac12 cdot EF cdot CE cdot sin angle CEFfrac12 cdot DE cdot BE cdot sin pembe BED fracEFDE cdot fracCEBE cdot fracsin angle CEFsin angle BED frac1996.mwishopanga 