a) Trong tam giác vuông ACD, ta có:
\(AC = \displaystyle {{AD} \over {\cos \widehat {CAD}}} = {{2,8} \over {\cos 74^\circ }}\)\( \approx 10,158\,(cm)\)
b) Kẻ \(DN \bot AC\)
Trong tam giác vuông \(AND\), ta có:
\(\eqalign{& DN = AD.\sin \widehat {DAN} \cr & = 2,8.\sin 74^\circ \approx 2,692\,(cm) \cr} \)
\(\eqalign{& AN = AD.\cos \widehat {DAN} \cr & = 2,8.\cos 74^\circ \approx 0,772\,(cm) \cr} \)Vì \(BX // DY\) nên \(\widehat {D{\rm{YX}}} = \widehat {BXY} = 123^\circ \) ( hai góc so le trong)
Mà \(\widehat {DYN} + \widehat {D{\rm{YX}}} = 180^\circ \) (kề bù)
Suy ra:
\(\widehat {DYN} = 180^\circ - \widehat {D{\rm{YX}}} = 180^\circ - 123^\circ\)\( = 57^\circ \)
Trong tam giác vuông \(DYN\), ta có:
\(\eqalign{& NY = DN.\cot \widehat {DYN} \cr & \approx 2,692.\cot 57^\circ \approx 1,748\,(cm) \cr} \)
Ta có:
\(\eqalign{& XY = AX - (AN + NY) \cr = 5,5 - (0,772 + 1,748) = 2,98cm \cr} \)
c) Ta có:
\(CX = AC - AX \approx 10,158 - 5,5\)\( = 4,658\,(cm)\)
Kẻ \(BM \bot CX\)
Ta có:
\(\widehat {BXC} = 180^\circ - \widehat {BXA} = 180^\circ - 123^\circ\)\( = 57^\circ \)
Trong tam giác vuông BMX, ta có:
\(\eqalign{& BM = BX.\sin \widehat {BXC} \cr & = 4,1.\sin 57^\circ \approx 3,439\,(cm) \cr} \)
\(\eqalign{& {S_{BCX}} = {1 \over 2}BM.CX \cr & = {1 \over 2}.3,439.4,658 = 8,009\,\left( {c{m^2}} \right). \cr} \)