题目 #19979

如图所示过点A作AF ⊥CD于点F，∵AF ⊥CD, CD ⊥BD, AB ⊥BD,所以∠AFD = ∠FDB = ∠B = 90∘, ∴ 四边形ABDF是矩形，所以AB = DF = 9m, AF = BD, AF ∥BD，∴∠FAE = ∠AEB = 60∘,所以在Rt △ ABE中，∠BAE = 30∘，∴AE = 2BE,\(AE^{2}\)=\(AB^{2}\)+\(BE^{2}\),所以（2BE）^{2}=\(9^{2}\)+\(BE^{2}\), 解得BE = 3√3m( 负值舍去)，所以BD = BE + DE = (3√3 + 30)m = AF, 在Rt △ACF中，因为tan∠CAF = tan35∘= _{AF}，所CF 以CF = AF ⋅tan35∘= (3√3 + 30)\(\times\)tan35∘≈(\(3 \times 1.7 + 30\))\(\times 0.7 = 24.57\)≈25m,所以CD = CF + DF = 25 + 9 = 34m, ∴ 楼房CD的高度为34m. 2.12米 如图,过点G作GH ⊥AB, 垂足为H. 由题意得GH = BF, GF = BH = 1米.设BC = x米, 所以FB = GH = FC − BC = (17 −x)米. 在Rt\(\Delta{}AGH\)中,∠AGH= 45∘, ∴AH = GH = (17 −x)米, ∴AB = AH + BH = (18 −x)米. 由题 意得∠ACB = ∠ECD, ∠ABC = ∠D = 90∘, 所以\(\Delta{}ABC\)∽\(\Delta{}EDC\)所以AB BC AB\(x _{ED}\)= _{DC}, ∴ _{1.8}= _{0.9}, 解得AB = 2x, ∴18 − x = 2x, 解得x = 6,所以AB = 2x = 12. 答: 古塔AB的高为12米. 3.(1)30∘; (2)(1200 −400√3)m. 如图，由题意可得∠CBE = 60∘, ∠CAF = 30∘, ∠BDM = 45∘, BM ⊥DM,BE ∥AF ∥DM．∴∠BCM = ∠CBE = 60∘, ∠ACM = ∠CAF = 30∘.所以∠ACB = ∠BCM −∠ACM = 60∘−30∘= 30∘； (2)∵∠CBE = 60∘，所以∠CBM = 90∘−∠CBE = 90∘−60∘= 30∘. 由(1)得∠ACB = 30∘．∴∠ABC=∠ACB= 30°, ∵AB = 800, ∴AB = AC = 800,在 Rt △ACM 中，sin∠ACM = AM\(CM _{AC}\),cos∠ACM= _{AC}, ∴AM = AC ⋅ sin∠ACM =\(800 \times\)sin30° =\(800 \times _{2}= 400\)，CM=AC ⋅coa∠ACM =\(800 \times\)cos30° =\(800 \times\sqrt{3_{2}}\)= 400√3，∴ 1 BM = BA +AM=800 + 400 = 1200, ∵∠BDN = 45°，BM ⊥DM, ∴DM = BM = 1200,所以DC = DM −CM = 1200 −400√3m, ∴景点C与景点D之间的距离为(1200 −400√3)m. 4.桂花树DF的高约为12.0m． 如图所示过点D作DG ⊥BP, 交BP延长线于点G，∴BG = AD,所以DG = AB = 25m, 在Rt △ABC中 , ∠BAC = 90∘, ∠ACB = ∠CBG = 55∘,且AB = 25 m, tan∠ACB = AB 25\(25 _{AC}\)，∴tan55∘= _{AC}, 即 _{AC}≈1.43,解得AC ≈ 17.48，∵CD = 8 m, ∴BG = AD = CD + AC ≈8 + 17.48 = 25.48(m)，因为DG ⊥BP, ∴∠G = 90∘, 在Rt △ BFG中, BG = 25.48 m, ∠FBG = 27∘,且tan∠FBG = _{BG}，∴tan27∘=\(FG _{25.48}\), 即 _{25.48}≈0.51, 解得FG ≈12.99， FG FG 所以DF = DG −FG = 25 −12.99 ≈12.0(m), 答; 桂花树DF的高约为12.0m． 5.6米. 如图,过点G作GN ⊥AB交AB于点N. 因为在Rt\(\Delta{}ANG\)中, ∠ANG = 90∘,所以tan53.13∘= AN\(4 _{NG}\)≈ _{3}, 设AN = 4a米, 则NG = 3a米, 因为AB ⊥CF,GF⊥CF, GN ⊥AB, ∴∠ABF = ∠CFB = ∠GNB = 90∘, ∴四边形BFGN为 矩形, ∴GF = GH + HF = 1.7 + 1.9 = 3.6米, ∴BF = NG = 3a米. 所以AB = (3.6 + 4a)米, ∴BE = 1.8 + 6 − 3a = (7.8 −3a)米, ∠DCE =∠ABE= 90∘, ∠E = ∠E, ∴\(\Delta{}DCE\)∼\(\Delta{}ABE\)解得a = 0.6，∴AB = 3.6 + 4a =3.6+\(4 \times 0.6 = 6\) (米). 答: 路灯AB的高度约为6米. 6.改造后的斜坡式自动扶梯AC的长度约为19.2米． 在Rt △ABD中，∠ABD = 30∘, AB = 10m，∴AD = ABsin∠ABD =\(10 \times\)sin30°= 5, 在Rt △ ACD中，∠ACD = 15∘, sin∠ACD = AD AD 5\(5 _{AC}\)，∴AC=sin∠ACD= _{sin15}∘≈ _{0.26}≈19.2m, 即：改造后的斜坡式自动扶梯 AC的长度约为19.2米． 7.珍珠女雕塑的总高度大约是5.8米． 如图设珍珠顶为点M, 小明站在C点处头顶为点D, 过点A作AE ∥BC交ME于点E, 则AE ⊥ME, 且点D在AE上, 由题意得AB = CD = 1.8米，且BC = AD = 4.6米, ∠MAE = 30∘, ∠MDE = 60∘, 设ME = x米,在Rt △AEM中, ∠MAE = 30∘，∴AE = ME x tan∠MAE= _{tan30}∘= √3x米,在Rt △MDE中, ∠MDE = 60∘，∴ DE = tan∠MDE= _{tan60}∘=\(\sqrt{3_{3}}\)x米，因为AE −DE = AD, ∴√3x −\(\sqrt{3_{3}}\)x = 4.6, 解得：x ≈3.984，所以总高度= ME x 3.9836 + 1.8 ≈5.8米, 珍珠女雕塑的总高度大约是5.8米． 8.(1)4m; (2)27∘. (1)过B作BE ⊥OC于E, 如图1所示: 在Rt\(\Delta{}ABE\)中, AB = 3m, ∠BAC = 53∘,因为sin∠BAC =\(BE _{AB}\), ∴BE = AB ⋅sin∠BAC =\(3 \times\)sin53∘≈\(3 \times 4 12 12 _{5}= _{5}(m)\).在直角三角形BOE中, ∠DOC = 37∘, BE = _{5}m, 12 因为 sin∠DOC = _{OB},所以OB =BE sin∠DOC=\(BE _{sin37}\)∘=BE\(\frac{5}{3} = 4(m)\).答: OB 的长约为 4m. 5 (2)过D作DF ⊥OC于F, 设云梯DO旋转后的对应点为P, 过点P作PG ⊥OC于点G, 过D作DH ⊥ PG于H, 如图2所示: ∴∠DFG = ∠PGF = ∠DHG = 90∘,PG-DF = 3(m), 所以四边形DFGH是矩形, ∠DOC = 37∘,因为sin∠DOC = DF\(3 _{OD}\), ∴DF = OD\(\times\)sin∠DOC =\(10\times\)sin37∘≈\(10 \times _{5}= 6(m)\)∵PG −DF = 3, ∴PG = 3 + DF = 3 + 6 =9(m),在Rt\(\Delta{}OGP\)中, PG = 9m, OP = OD = 10m, ∴∠DFG = ∠PGF =∠DHG= 90∘, PG −DF = PG 9 3. ∵sin∠POG = _{OP}= _{10}= 0.9, ∴∠POG ≈64∘,所以∠POD = ∠POG −∠DOC ≈64∘−37∘= 27∘.答： 云梯OD旋转的度数约为27∘. 9.河流CD的宽度为220米． 过点B作BN ⊥MD于点N，如图所示：则∠BNM = ∠BND = 90∘，由题意可得：AF ∥MD, ∠BAM = ∠AMN = 90∘, AB = 60米，∴四边形AMNB为矩形，∴MN = AB = 60米,BN = AM, ∴NC = MC − MN=80−60 = 20(米), ∵AF ∥MD, ∴∠ACM = ∠FAC = 60∘, ∠BDC =∠FBD= 30∘, 在Rt △ACM中，AM = CM ⋅tan∠ACM = MC ⋅tan60∘= 80√3米，∴BN = AM = 80√3米, ∴ND = BN 80√3 80√3 tan∠BDC= _{tan30}∘= √3 = 3 240(米),∴CD = ND −NC = 240 −20 = 220(米).答：河流CD的宽度为220米． 10.该铁塔的高AE为58米． 如图过点C作CF ⊥AB于点F, 设塔高AE = x, 作CF ⊥AB于点F,则四边形BDCF是矩形，∴CD = BF = 30m, CF = BD, 在Rt △ADB中，因为∠ADB = 45∘，∴AB = BD = x + 62, 在Rt △ACF中, ∠ACF = 36∘52′, 且CF = BD = x + 62, AF = x + 62 −30 = x + 32，∴tan36∘52′= x+\(32 _{x+62}\)≈0.75,所以x = 58. 答; 该铁塔的高AE为5