题目 #19794

(1)①如图所示, 延长OA交BC于F, ∵BC ∥OE, OA ⊥OE, ∴OF ⊥BC,所以∠AFB = 90∘, ∴∠BAO = ∠B + ∠AFB = 160∘, 故答案为: 160. ②在Rt △ABF中, AB = 40cm, ∠B = 70∘, ∠AFB = 90∘, 所以AF = AB\(\times\)sinB≈\(40 \times 0.94 = 37.6cm\)∴ OF = OA + AF = 44cm, 因为CD = 8cm,所以投影探头的端点D到桌面OE的距离为44 −8 = 36cm, 故答案为: 36. (2)如图所示, 延长CD交OE于点H, 过点B作BM ⊥CD, 与DC延长线相交于M, 过A作AF ⊥BM于F, 则四边形FMHO是矩形, 由(1)得∠MBA = 70∘,OF= 44cm ∴MH = OF = 44 cm, 因为∠ABC = 30∘, 所以∠MBC = 40∘,在Rt\(\Delta{}MBC\)中, BC = 45cm, ∠M = 90∘, 所以CM = BC ⋅sin∠MBC ≈\(45 \times 0.64= 28.8cm\)，∴DH = MH −CD −CM = 7.2cm, ∴投影探头的端点D到桌面OE的距离为7.2cm. 2.1.8m 如图:延长CE交BM于G, 因为AM ⊥DM, CD ⊥DM, EF ⊥DM, CD =EF= 1.6m, 所以四边形EFMG为矩形 , 所以EG = FM = 13.4m, CG =DM= 15.2m, GM = EF = CD = 1.6m, 所以BG = BM −GM = 15m −1.6m= 13.4m, 所以BG = EG. 又因为CG ⊥BM, 所以∠2 + ∠EAB = 90∘.因为∠1 + ∠2 = 90∘, ∴∠1 = ∠1 = ∠EAB ∠EAB,在\(\Delta{}CGB\)和\(\Delta{}AGE\)中 {∠AGE = ∠BGC,所以\(\Delta{}CGB\)≅\(\Delta{}AGE(AAS)\)所以 AG = CG = 15.2m, ∴AB = BG = EG AG −BG= 15.2m −13.4m = 1.8m. ∴宣传牌AB的高为1.8m. 2026-06-22