题目 #19669

设AB, EF交于D, ∵∠DAF = 30∘, ∴∠ADF = 90∘−30∘= 60∘, ∴∠BDE等于60∘, 在Rt △BDE中, sin∠BDE = _{DE}, ∴\(\sqrt{3_{DE}}\)=\(\sqrt{3_{2}}\), 解得: DE = 2(m),所以BD = 1(m), ∴AD = AB −BD = 5(m), 在Rt △ADF中, ∠DAF = 30∘,所 BE 以DF = 1\(5 _{2}\)AD = _{2}(m). ∴EF = DE + DF = _{2}(m), 故答案为: _{2}.^{9} 9 8.63米 如图:作 BN ⊥CD 于 N, BM ⊥AC 于 M,在 Rt\(\Delta{}BDN\)中,因为 tan∠D = BN\(1 _{DN}\)= _{2}, BD = 10√5,设BN = x, 则DN = 2x，在Rt\(\Delta{}BND\)中, 由勾股定理可得:\(BN^{2}\)+\(DN^{2}\)=\(BD^{2}\),即\(x^{2}\)+ (2x)^{2}= (10√5)^{2}, 解得: x = 10或x = −10(负数舍去), ∴BN = 10, DN = 20,因为 ∠C = ∠CMB = ∠CNB = 90°, 所以四边形CMBN是矩形,所以 CM = BN = 10, BM = CN = CD −DN = 60 −20 = 40, 在Rt\(\Delta{}ABM\)中, tan∠ABM = tan53∘=\(AM _{BM}\)≈1.327, 所以AM ≈53.08,所以 AC = AM + CM ≈53.08 + 10 ≈63(米), 答: 建筑物AC的高度约为 63米． 9.楼房高度约为23米 过点D作DE ⊥AB于点E，过点C作CF ⊥DE于点F，由题意得∠1 = 45∘, ∠2 = 30∘, DE = 40m, ∴∠3 = 90∘−∠1 = 45∘, ∠4 = 90∘−∠2 = 60∘因为DE ⊥AB, CF ⊥DE, ∴∠FEB = ∠CBE = ∠CFE = 90∘, 所以 四边形FEBC为矩形，∴BC = FE, CF = BE, ∵在Rt △ADE中，AE = DE\(\times\)tan∠3 = DE\(\times\)tan45∘= DE = 40m，所以BE = AB −AE = 70 −40 = 30 m, ∴CF = BE = 30 m ∵ 在 Rt △CFD中, DF = CF CF tan∠4= _{tan60}∘=\(30 \times\sqrt{3_{3}}\)= 10√3m ∴BC = EF = DE −DF等于40 −10√3 ≈23m答：楼房高度约为23m． 10.6 在Rt △ABC中, ∵∠ACB = 90∘, ∴BC = AB ⋅sin∠BAC =\(12 \times 1 _{2}= 6\)(米). 11.33米. 过点E作EH ⊥BH, 过点E作EW ⊥AC, 因为EH ⊥BH, EW ⊥AC,所以∠EHA = ∠A = ∠EWA = 90°, 所以四边形 EHAW 是矩形,所以 EH = AW, HA = EW,因为斜坡 BE 下方标示 BE 的坡度 i = 1: 2.5,所以\(EH _{BH}\)= _{2.5}, 即BH = 2.5EH,在Rt\(\Delta{}EHB\)中,\(EB^{2}\)=\(EH^{2}\)+\(HB^{2}\), 因为斜坡BE的长度为3√29米,即\(9 \times 29 =EH^{2}\)+ 1\(HB^{2}\)= 7.\(25EH^{2}\), 解得EH = 6(负值已舍去),HB =\(2.5 \times 6 = 15\)(米)∴设AB = r, 则WE = HA = 15 + r, 在 Rt\(\Delta{}CEW\)中,tan∠CEW = CW\(CW _{EW}\), 即tan35° = _{15+r}, ∴CW = 0.7(15 + r),所以AC = CW + AW = CW + EH = 0.7(15 + r) + 6 = 16.5 + 0.7r,在 Rt\(\Delta{}CAB\)中,tan∠ABC = AC\(CA _{AB}\), 即tan54° = _{r}, ∴CA = 1.4r, 则1.4r = 16.5 + 0.7r,所以0.7r = 16.5,解得 r = 165\(165 _{7}\), ∴CA =\(1.4 \times _{7}= 33\)米. 12.都梁阁的高度为48米 如图所示, 延长AB交CE的延长线于点F, 则AF ⊥CF, 根据题意, CD = 90m, CD ⊥AD, AB ⊥AD, ∵在Rt △ BEF中, ∠BEF = 45∘, ∴BF = EF,因为在Rt △CDE中，∠CED = 45∘, CD = 90 m，∴CE = CD = 90 m,所以 在Rt △CFB中，∠BCF = 18∘，令BF = x，∴tan∠BCF = BF\(x _{CF}\),所以tan18∘= _{90+x}≈0.32, ∴x = 42.3, ∴ BF = 42.3 m, ∵CD = AF,所以AB + BF = 90即AB + 42.3 = 90，解得AB = 47.7 ≈48( m)，答 ：都梁阁AB的高度为48m． 13.加高后的坝底HD的长为29.4m． 由题意得 BG = 3.2m, MN = EF = 3.2 + 2 = 5.2m, ME = NF = BC = 6m,因为在 Rt △DEF 中， EF\(1 _{FD}\)= _{2}，∴ FD = 2EF =\(2 \times 5.2 = 10.4\)m,因为在 Rt △HMN 中，所以MN\(1 _{HN}\)= _{2.5}, ∴HN = 2.5MN =\(2.5 \times 5.2 = 13m\)，∴ HD = HN + NF + FD=13+6+10.4= 29.4m, 答; 加高后的坝底HD的长为29.4m． 14.钢支架AB的高度约为18.8米. 如图所示,过点 E 作 EF ⊥CD 于 F, ∵DE 的坡度为 3: 4, ∴ EF\(3 _{DF}\)= _{4},设EF = 3x米, DF = 4x米, 在Rt\(\Delta{}DEF\)中, 由勾股定理得\(DE^{2}\)=\(EF^{2}\)+\(DF^{2}\),所以\(50^{2}\)= (3x)^{2}+ (4x)^{2}, 解得x = 10或x = −10(舍去), ∴EF = 30米 , DF = 40米;因为BE ⊥AB, AC ⊥CD, EF ⊥CD, ∴四边形EFCB是矩形,所以BE = CF, BC = EF = 30米, 设 AB = y 米,则 AC = (y + 30)米,在Rt\(\Delta{}ABE\)中, BE = AB y tan∠AEB= tan65°≈0.467y米, 在Rt\(\Delta{}ADC\)中,CD = AC y+30 tan∠ADC= tan45°= y + 30米, ∴0.467y + 40 = y + 30,解得y ≈18.8米, 答: 钢支架AB的高度约为18.8米. 15.乙楼的高为39m 过点A作AC ⊥BD, 如图, 由题意得，∠AED = ∠EDC = ∠ACD = 90°, 所以四边形AEDC为矩形, 所以CD = AE = 18m, AC = DE = 30m, 依题意得∠BAC = 35°, 所以∠ACB = 180°−∠ACD = 90°, 因为 在Rt △ABC中，tan∠BAC = BC AC , 所以BC = ACan∠BAC = 21m, 所以 BD = BC + CD = 39m, 所以乙楼的高为39m.