题目 #19663

∵BC ⊥AC, ∠A =\(\alpha\)AC = 10米, ∴BC = AC ⋅tan\(\alpha = 10tan\alpha\). 故选A. 2.109.2米 如图,延长AE交CD于点M, 过点A作AN ⊥BC于点N, 由题意得,得∠AMC = ∠NCM = ∠ANC = 90°, ∴ 四边形AMCN为矩形,所以NC = AM, NA = CM, 在Rt\(\Delta{}EMD\)中, ∠EMD = 90°,所以sin∠EDM = EM DM EM DM 3\(4 _{ED}\), cos∠EDM = _{ED}, 即sin37° = _{30}, cos37° = _{30},所以EM ≈\(30 \times _{5}= 18\)DM ≈\(30 \times _{5}= 24\)又tan42.6° = BN\(BN _{AN}\)= _{74+24},所以BN =\(98 \times\)tan42.6° ≈88.2, ∴BC = BN + AE + EM = 88.2 + 3 + 18 = 109.2,故大 楼BC的高度约为109.2米. 3.楼房AB的高度为(20 + 8√3)米. 过点E 作 EF ⊥BC 的延长线于点 F, EH ⊥AB 于点 H.在 Rt △CEF 中，因为i = EF\(1 _{CF}\)= √3= tan∠ECF，∴ ∠ECF = 30∘∴EF =\(1 _{2}\)CE = 10米, CF = 10√3米所以BH = EF = 10米，HE = BF = BC + CF = (24 + 10√3) 米在Rt △AHE中∵∠AEH = 30∘, ∴tan30∘=\(AH_{HE}\)AH等于HE ⋅tan30∘= (24 + 10√3)\(\times\sqrt{3_{3}}\)= (8√3 + 10) 米∴ AB等于AH + HB = (8√3 + 10) + 10 = (20 + 8√3)米, 故楼房AB的高为(20 + 8√3)米 4.B 已知在直角三角形 ABC 中, ∠B = 90°, ∠37°,BC=20m,可得 tanC=\(AB_{AC}\),则AB = BC ⋅tanC = 20tan37°. 故选 : B. 5.(1)15°; (2)海监船继续向正东方向航行安全. (1)如图, 过点P作PD ⊥AB于点D, 由题意得, ∠PAB = 30°, ∠PBD = 45°,所以∠APB = ∠PBD −∠PAB = 15°, 故∠APB的度数为15°; (2)∵海监船以每小时40海里的速度向正东方向航行,在A处测得灯塔P在北偏东60°方向上, 继续航行 30 分钟后到达 B 处, ∴AB =\(40 \times 1 _{2}= 20\)(海里)设BD = x海里, ∴AD = (20 + x)海里, ∵∠PBD = 45°, ∠PDB = 90°, ∴∠BPD = 45°,所以PD = BD = x 海里, ∵∠PAB = 30°, ∴tan30° = PD\(x _{AD}\)= _{x+20}=\(\sqrt{3_{3}}\), 解得 : x = 10√3 + 10 ≈27.32 > 25, 即海监船继续向正东方向航行安全. 6.14.3 如图所示, 过点P作PE ⊥AB, 垂足为E, 过点P作PF ⊥BD, 垂足为F,所以∠PEB = ∠EBF = ∠PFB = 90∘, ∴ 四边形 BEPF 是矩形，所以PF = BE, PE = BF, ∵ 山坡坡度 i = 5: 12， ∴ PF\(5 _{CF}\)= _{12}, 设PF = 5k, CF = 12k,在 Rt △ABC中, ∠ACB = 63.4∘, BC = 90米, 所以AB = BC ⋅tan63.4∘= 180 (米),AE = AB −BE = (180 − 5k) 米,EP = BF = BC + CF = (90 + 12k)米，在,在Rt △AEP中，∠APE = 53∘，∴tan53∘= AE 180−\(5k _{EP}\)= _{90+12k}= 4 20 20\(100 _{3}\), ∴k = _{7},经检验，k = _{7}是原方程的根，∴PF = 5k = _{7}≈14.3 (米), ∴此人所在 位置点P的铅直高度为14.3米． 9 7. 2 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 =\(2.5 \times 5.2 = 13m\)