1) cos<A =cos(180° -<B) = - cos<B = - (-1/3) =1/3. 2) AD|| BC ;AD =13; BC = 5 ; cosA .=0,8 .S =1/2*(AD +BC)* h , где h высота трапецииПроведем BH ┴ AD . AH = (AD -BC)/2 =(13 -5)/2 =4 ;Из ΔAHB: h =BH =AH*tqA ;sinA =√(1 -cos²A) =√(1 -O,8²) =0,6;tqA =sinA/cosA =0,6/0.8 =3/4;h =4*3/4 =3.S =1/2(13+5)*3 =27.3) 1 + ctq²A = 1/sin²A ; ;sin²A = 1/(1+ctq²A);sin²A = 1/(1+1/tq²A);sin²A = 1/(1+1/5²) ;sin²A =25/26 ;sinA =5/√26 =5√26/26.