数学成长资源(Mathematical growth resources)

数学成长资源（Mathematical growth resources）26.2 see a quadratic equation with the view of a functionSmart upgrade1. A2. B3. D4. A5. D6. C7.6 8. Minus 2 is less than or equal to 1(1) y = (x - 1) 2-1 (2) - 1 or 3(3) less than 0 or greater than 2 is greater than 0 and less than 210. Solution: y = x2-2x - 1 = x2-2x + 1-1-1 = (x-1) 2-2,∴ vertex coordinates of (1, 2).So y is equal to 0, x2 minus 2x minus 1 is equal to 0.X1 is equal to 1 plus 2, x2 is equal to 1 minus 2.∴ and intersection of the x a xis for the (1 + 2, 0), (1-2, 0).(1) substitute (0, 3) into y = -x2 + (m - 1), and m = 4.(2) minus x2 plus 3 is equal to 0, so x is equal to plus or minus 3,∴ and intersection of the x axis (3, 0), (3, 0), the vertex coordinates of (0, 3).(3) -3 < x < 3, the parabola is above the X-axis.(4) when x > 0, the value of y decreases with the increase of x value.The solution: (1) x2-2x - 3 = 0, x1 = -1, x2 = 3.∴ A (1, 0), B (3, 0).So if you plug in the coordinates of A, B, and B, and then you plug in y = ax2 + bx + 2,So you get a minus b plus 2 is equal to 0, 9, a plus 3b plus 2 is equal to 0.(2) it has to be (1) y = -23x2 + 43x + 2.∵ when x = 0, y = 2, ∴ C (0, 2).Let's say that the analytic formula for line AC is y is equal to k x plus b, and I'm going to put the two coordinates of A and C at y is equal to kx plus b, so k is equal to 2, and b is equal to 2.∴ analytical type of linear AC y = 2 x + 2.Similarly, the analytical formula for the line BC can be obtainedY is equal to minus 23x plus 2.Direct tests1. C2. D3. X = -1,4.12.55.4(1) if you want to make a point, you can get 22 + 2p + + 1 = 0.So q is equal to minus 2p plus 5.(2) to prove: a yuan quadratic equation ∵ x2 + p + q = 0 discriminant Δ = p2-4 q,By (1)Δ = p2 + 4 (2 p + 5) = p + 20 p2 + 8 = (2 + 4 p + 4) > 0,∴ a yuan quadratic equation x2 + p + q = 0, there are two not equal to the real root.∴ parabolic y = x2 + p + q has two intersections with the x axis.7. Solution: (1) drawing as shown.Depends on the questionY is equal to x minus 1, 2 minus 2 is x2 minus 2x plus 1 minus 2 is x2 minus 2x minus 1.∴ translation after image analytic expression for y = x2-2-1 x.(2) when y is 0, x2 minus 2x minus 1 is 0,(x - 1) 2 = 2,X minus 1 is equal to plus or minus 2,X1 is equal to 1 minus 2, x2 is equal to 1 plus 2.∴ the translation image and x axis to two points, coordinates, respectively (1-2, 0) and (1 + 2, 0).It can be seen from the diagram that when x < 1-2 or x > 1 + 2, the quadratic function y = (x - 1) 2-2 is greater than 0.26.3 practical problems and quadratic functionsSmart upgrade1. D2. D3. Y = -30x + 960 w = -30x2 + 960x - 14404. S = 2x25. Solution: (1) the profit of each loaf is (x-5) Angle, and the number of bread sold is (300-20x) or [160 - (x-7) x 20].(2) y = (300-20x) (x-5) = -20x2 + 400x - 1500,So y is equal to minus 20x2 plus 400x minus 1500.(3) y = -20x2 + 400x - 1500 = -20 (x-10) 2 + 500.∴ when x = 10, a maximum of 500 y.∴ when each bread unit price as the Angle of 10, the retail profits biggest daily, Angle of maximum profit of 500.6. Solution: (1) 17.38(2) after drinking, when v = 17, s = 46,Plug in s = TV + 0.08v2, you get t is equal to 1.35.If the speed of drinking is 11 m/s, the brake distance is s = 1.35 * 11 + 0.08 * 112 = 24.53. The brake distance of the non-alcoholic drink is 17.38, so it increases by 24.53-17.387.2 m.(3) it is understood that 17t + 0.08 * 172 < 40, solution t < 0.99. Your reaction time should be no more than 0.99 seconds.7. Solution: (1) y = (x-20)? W is equal to x minus 20 times minus 2x plus 80.= -2x2 + 120x - 1 600,∴ function relation between the y and x is y = 120-2 x2 + x - 1, 600.(2) y = -2x2 + 120x - 1600 = -2 (x-30) 2 + 200,∴ when = 30 x, y has a maximum of 200.When the selling price is RMB 30 / kg, the maximum profit is 200 yuan per day.(3) when y = 150, the equation is equal to 2 (x-30) 2 + 200 = 150.So if you solve this equation, you get x1 is equal to 25, x2 is equal to 35.According to the question, x2 = 35 is not an issue.∴ when the sale price to $25 / kg, the farmers get the sales profit is 150 yuan per day.X: (1) y = kx + b, by image30k + b = 400,40k + b = 200. The solution is k = -20, b = 1 000.∴ y = 000-20 x + 1 (30 x 50 or less or less).(2) P = (x-20) y = (x-20) (-20x + 1 000)= -20x2 + 1 400x - 20 000.∵ - 20 < 0, ∴ P has a maxim um value.When x is minus 1, 400, 2 times? - 20? = 35,P maximum = 4 500.The maximum profit is 4 500 yuan per day when the unit price is 35 yuan per kilogram.(3) 31 x or less acuities were 34 or 36 x or less 39 or less.Direct tests1.252(1) according to the meaning, we have toY = (2, 400-2, 000 - x) (8 + 4 x x50),So y is equal to minus 225x2 plus 24x plus 3, 200.(2) by means of a questionMinus 225x2 plus 24x plus 3, 200 is equal to 4,800.So you get x2 minus 300, x plus 20, 000 is equal to 0.So to solve this equation, x1 is equal to 100, x2 is equal to 200.In order to make the people affordable, x = 200. Therefore, every refrigerator should be reduced by 200 yuan.(3) for y = -225x2 + 24x + 3, 200,When x is equal to minus 242 times? - 225? = 150,The maximum value of y = (2, 400-2, 000-150) (8 + 4 x 15050).= 250 times 20 is 5,000.Therefore, when the price of each refrigerator is 150 yuan, the profit of the mall is the largest, and the maximum profit is 5,000 yuan.Chapter xxvii similar to 27.1 figuresSmart upgradeB 1. A2. D3. C4.5. 2.6.10, 14,7.400 m8. Solution: no, because the length and width of glass similar ratio of 26:18 = 13:9, and frames the length and width ratio of 26 + (4) : (18 + 4) = 15:11 indicates: 9, 13 so glass and frame is not similar to that of the rectangular outside.9. Solution: the shape of A quadrilateral ABCD is similar to the quadrilateral A 'B' C 'D< A = < A '= 150 °,< D = 360 ° 150 ° (+ 60 °, 75 °) = 75 °.B 'C'C 'C' C 'B' AB = C 'D' CD,B 'C' 8 = 2.52 =C 'D' 5,B 'C' = 10,C 'D' = 254.10. Solution: the line segment given is: 22 cm, 2 cm, etc. Because 12 is 22, 2, 12 is 22.The answer is not unique, as long as it is proportional. Direct tests1. D2. B3. C4. A.Because = < < AB C DEF = 45 ° 90 ° + = 90 °,ABDE = 22 = 2, BCEF = 222 = 2,∴ ABDE = BCEF.∴ delta ABC ∽ delta DEF.When the BD = a2b, delta ACB ∽ delta CBD.7. Solution: (1) the ∵ < ACP = < PDB = 120 °,When ACPD = PCDB, ACCD = CDDB,CD2 is equal to AC, right? The DB, delta ACP ∽ delta PDB.(2) ∵ delta ACP ∽ delta PDB, ∴ < A = < DPB.∴ < APB = < APC + + < < CPD DPBSo this is Angle APC plus Angle A plus Angle CPD= < PCD + < CPD = 120 °.8. Solution: because point D, I is the third point of AB and CA,∴ ADAB = AIAC = 13.And ∵ < A = < A, ∴ delta ADI ∽ delta ABC.∴ DIBC = AIAC = 13, DI = 13 BC.The same thing is EF = 13AC, GH = 13AB.∵ point D, E, F, G, H, I were third point of the AB, BC, CA,∴ DE = 13 ab, FG = 13 BC, HI = 13 ac.∴ hexagon DEFGHI perimete r= DE + EF + FG + GH + HI + ID= 13AB + 13AC + 13BC + 13b + 13AC + 13BC= 23 (AB + AC + BC) = 23a.9. Solution: set AP = x, and BP = 6 - x.∵ AD ∥ BC, B < = 90 °, ∴ < A = 90 °.∴ < A < = B.(1) when APBP = ADBC, delta APD ∽ delta BPC,∴ x6 - x = 18, ∴ x = 23.(2) when APBC = ADBP,Delta APD ∽ delta BCP,∴ by 8 = 16 - x.∴ = 2 x, or x = 4.∴ the desires of AP long 23, 2 or 4.Direct tests1. A2.3Angle ACD = Angle B (Angle ADC = Angle ACB or ADAC = ACAB)4. Solution: (1) to prove: ∵ delta A BC is an equilateral triangle.∴ < BAC = < ACB = 60 °, < ACF = 120 °.∵ CE is exterior Angle bisector,∴ < ACE = 60 °.∴ < BAC = < ECD.And ∵ = < < the ADB CDE,∴ delta ABD ∽ delta CFD.(2) BM orthogonal complement AC at point M, AC = AB = 6.∴ AM = CM = 3, BM = AB? Sin60 ° = 33.∵ AD = 2 CD,∴ CD = 2, AD = 4, MD = 1.In Rt delta in BDMBD = BM2 + MD2 = 27.By (1) delta ABD ∽ delta CED,BDED = ADCD, 27ED = 2.∴ ED = 7.∴ BE = BD + ED = 37.27.2.2 examples of similar triangles Smart upgrade1. C2.7.53.10 m4. Solution: ∵ AB an BD, AC an AB,∴ AC ∥ BD.∴ < ACB = < DBC.And ∵ < A = < BCD = 90 °,∴ delta ABC ∽ delta CDB.∴ ACBC = BCBD.∴ BC2 = AC? BD.In Rt delta ABC,BC2 = AC2 + AB2 = 152 + 302 = 1 125, ∴ AC? BD = 1 125∴ 15? BD = 1, 125.∴ BD = 75 (cm).5. Prove that: (1) the ∵ quadrilateral ABCD and quadrilateral DEFG are square.∴ AD = CD, DE = DG,ADC = < < EDG = 90 °,Angle ADC + Angle ADG = Angle EDG + Angle ADGThe < ADE = < the CDG.∴ delta ADE ≌ delta CDG. (SAS)∴ A E = CG.(2) by (1) to know delta ADE ≌ delta CDG,∴ < DAE will work = < DCG.And < ANM = < CND, ∴ delta AMN ∽ delta CDN.∴ ANCN = MNDN,Namely the AN? DN = CN? MN.6. Solution: (1) to prove: ∵ ABCD is a square.∴ < DAE will work = < FBE = 90 °.∴ <ADE + < DEA = 90 °.EF DE coming again,∴ < AED + < FEB = 90 °.∴ < ADE = < FEB. ∴ delta ADE ∽ delta BEF.(2) by (1) delta ADE ∽ delta BEF, AD = 4, BE = 4 - x,So yx is equal to 4 minus x4.So y is equal to 14 minus x2 plus 4x is equal to 14 times x minus 2 plus 4.It's minus 4 times x minus 2 plus 1.∴ when = 2 x, y has the maximum value, maximum of y is 1.7. Solution: ∵ CD an FB, AB an FB, ∴ CD ∥ AB.∴ delta CGE ∽ delta AHE.∴ CGAH = EGEH that CD - EFAH = FDFD + BD.∴ 3-1.6 AH = 22 + 15.∴ AH = 11.9.∴ AB + HB = = AH AH + EF = 11.9 + 1.6 = 13.5 (m).8. Solution: (1) the nature of the light and shadow is DE ∥ AC,∴ < BDE = < BAC, < BED = < BCA.∴ delta BDE ∽ delta BAC.∴ DEBD = ACAB.AC = 302 + 302 = 50 ∵ (m), BD = 83 (m),AB = 40 (m), ∴ DE = 103 (m).(2) BE = de2-bd2 = 2.The time spent by wang gang at E point is 40 + 23 = 14 (s), and zhang hua's time at the D point is 14-4 = 10 (s), and zhang hua's speed of chasing wang is (40-83), which is 10 (m/s).Direct tests1. B2. C3.9B: ABA is a good place to goAB = kA 'B', AC = kA 'C',In Rt delta ABC and Rt delta A 'B' C ',BCB 'C' = AB2 - AC2A 'B' 2 - A '2 - A' B '-' B '-' B '-' B '-' 2 - A '-' B '-' B '-' B '.∴ ABA 'B' = ACA 'C' = BCB 'C'.∴ Rt delta ABC ∽ Rt delta A 'B' C '.Answer: (1) an acute Angle corresponding to the corresponding ratio of two right angles(2) the hypotenuse is proportional to a right AngleIn Rt delta ABC and Rt delta A 'B' C ',< = C < C '= 90 °, ABA' B '= ACA' C '27.2.3 the perimeter and area of a similar triangleSmart upgrade1. C2. C3. B4. B5. D6.203 cm27.506.6498.1:69.1210. Solution: ∵ EF ∥ AB, ∴ delta ECF ∽ delta ACB.∴ S delta ECFS delta ACB = CE2CA2.When the area of delta ECF is equal to the area of thequadrilateral EABF,Delta ECFS delta, delta ACB = CE2CA2 = 12.∴ CE = 22 ca = 22.Solution: the area of a small triangle is xcm2, and 32102 = x400.X is equal to 36.The perimeter of the two triangles is 3k and 10k respectively10k minus 3k is equal to 560.So 3k = 240, 10, k = 800.The area of the small triangle is 36 cm2 and the circumference is 240 cm.The circumference of the large triangle is 800 cm.Direct tests1. B2.1:23.94.255. Solution: (1) to prove: ∵ DC to AC,∴ delta ACD is isosceles triangle.∵ CF split < ACD,∴ F is the halfway point of the AD.∵ E as the midpoint of AB,∴ EF for delta ABD the median line.∴ EF ∥ BC.(2) by (1) the EF ∥ BC, ∴ EFBD = 12.∴ S delta AEF: S train ABD = 1:4.∴ S quadrilateral BDEF: S delta ABD = 3, 4.∵ S delta ABD = 6,∴ S quadrilateral BDEF = 92.27.3 likelihoodSmart upgradeB 1. C2. D3.() 16. Solution:Figure (1), (2), (3), and (4) are the centers of O1, O2, O3, and O4 respectively.7.Solution: ∵ quadrilateral ABCD and quadrilateral EFGH is a graphic,∴ parallelogram ABCD ∽ quadrilateral EFGH.∵ parallelogram ABCD and the area of the quadrilateral EFGH ratio of 4, 9,∴ parallelogram ABCD and quadrilateral EFGH simil arity ratio of 2:3.Similar ∵ polygon circumference ratio is equal to the similar ratio,∴ parallelogram ABCD and quadrilateral EFGH ratio of the circumference of the 2, 3.Solution: as shown below.9. Solution: (1) to make a few pairs of lines CC ', BB 'and AA' in the corresponding points respectively, the three lines intersect at point O, and O is the desired.(2) as can be seen from the figure, A 'B' = 42 + 62 = 213,AB is equal to 22 plus 32 is equal to 13.∵ again A 'B' and AB is corresponding t o the side,A 'B: AB = 13:213 = 2:1,∴ delta ABC to delta A 'B' C 'of A ratio of 2:1.(3) as shown below.10. Solution: (1) (2) as shown.(3) the center coordinates (0, 0).(4) is the axisymmetric figure.Direct tests1. (1) draw the origin O, x axis, Y-axis. B (2, 1).(2) graph delta A 'B' C '.3 S is equal to 12 times4 times 8 is 16.Solution: figure.Chapter xxviii acute triangle function 28.1 acute triangle functionClass 1Smart upgrade1. C2. B3. A4. A.5.12136.457. (1), 3212 (2), 889898. Solution: ∵ AB = BC2 + AC2 = 289 = 17.∴ sin = BCAB = 817 A.9. Solution: c = a2 + b2 =? 63? 2 plus 182 is equal to 123,Minimum Angle of < a. ∴ sin A = 12.10. Solution: there are three pairs of proportional line segments of the sine of Angle B,Sine of B is equal to ACAB = CDBC = ADAC.Direct tests1. B2. A3.30 °4.555.356. Solution: as shown in the picturePoint B is the BC vertical bank, the perpendicular is C, and in Rt delta ACB, there isAB = BCsin < BAC sin60 = 900 °= 6003,So time t = sv = 60035 x 60= 2, 3, material (3.4 min).A: it takes about 3.4 minutes from A to B.The second classSmart upgrade1. D2. B3. C4. A5. D6.7247. (1), 810 (2), 5233638. 1313 (2), 33232 (1)(3), 21313413139. Solution: (1) sine of A = 35, cosine of A = 45, sine of B = 45, cosine of B = 35.(2) sin A = 817, cos A = 1517, sine B = 1517, cos B = 817.(3) sine of A = 158, cosine of A = 78, sine of B = 78, cosine of B = 158.10. Solution: (1) by BC = 6, AC = 8, according to the Pythagorean theorem,AB = BC2 + AC2 = 62 + 82 = 10∴ sinA = BCAB = 610 = 35(2) ∵ < ACB = 90 °, CD an AB∴ < A + A + < ACD = < < B∴ < ACD = < B∴ cos < ACD = cosB = BCAB = 610 = 35(3) ∵ S delta ABC = 12 ab? CD = 12 BC? AC∴ AB? CD = BC? AC∴ CD = BC? ACAB is equal to 6 times 810 is equal to 245Direct tests1. C2. B3. B4. A5. C6.45The third classSmart upgrade1. D2. B3. C4. B5. D6. > >7.1230 ° + 8. Solution: (1) the sin sin 60 ° 45 ° = 12-2 cos + 32-2? 22 is 12 times 3 minus 1.(2) 1-45 ° cos2-1-60 ° sin2 = 1 -? 22? 2 minus 1 --? 32? 2= 12 (2-1).(3) | sin 30 ° to 30 ° - cos | = 12-32 = 12 (3-1).(4) cos 45 ° to 45 ° sine and cosine 60 ° 1 + 30 ° 30 ° - 3 tan sin = 1-121 + 12-3 x = 23 33-3.9. Solution: c = a2 + b2 = 72 + 216 = 288 = 122.∵ tan A = ab = 6266 = 33,∴ < 30 °, A = < B = 90 ° - < = 60 °.∴ < 30 °, A = < B = 60 °, c = 122.10. Solution: ∵ | | + cos A - 22 (sin - 12 B) = 0, 2∴ | | cos A - 22 acuity 0, (sin - 12 B) 2 0 or higher,∴ | | cos A - 22 = 0,(sin B minus 12) 2 is equal to 0.∴ cos A - 22 = 0, sin - 12 B = 0,So cosine of A is equal to 22, sine of B is equal to 12.∴ < 45 °, A = B < = 30 °∴ < = 180 ° to 45 ° to 30 ° C = 105 °.Direct tests1. B2.333.24.55. Resolution: the length of the carpet should be determined. The total length of the carpet should be the sum of the line segment AC and BC.Because AB = 4 m, < BAC = 30 °, < = 90 ° C,SinA = sin30 ° = BCAB = 12, so the BC = 2 m, according to the Pythagorean theorem can be:AC = 23 m.So the length of the carpet on the AB section should beM (2 + 23).Answer: (2 + 23) m6. Solution: (1) in Rt delta ABC, sinB = 55, AB = 25, ACAB = 55.∴ AC = 2, according to the Pythagorean theorem: the BC = 4.(2) the ∵ PD ∥ AB,∴ delta ABC ∽ delta DPC.∴ DCPC = ACBC = 12.Set PC = x, DC = 12x, AD = 2-12x,∴ S delta ADP = 12 AD? PC = 12 (2-12x)? x= -14x2 + x = -14 (x-2) 2 + 1.∴ when = 2 x, y, a maximum of 1.Class 4Smart upgradeB 1. C2. C3.126.0.922 4.345.3 + 57. Solution: (1) the original form = -4 + 33 + 1-3 * (3-1) = 0.(2) the original equation is 1 + 3-2 times 12 = 4-1 = 3.(3) primitive = (22) 2-33 + 12 + 6 * 33So it's 12 minus 33 plus 12 plus 23 is equal to 1 minus 3.(4) the original formula is 3-1 + 12-1 + 12 = 3-1.(1) sine A = 0.8683, cosine A = 0.4962, tan A = 1.75.(2) < = 19.18 ° A, B < = 84.33 °.9. Solution: the original is equal to x plus 1x2 plus x minus 2? X + 2? X + 1?? X - 1? = 1? X - 1? 2.When x = tan 30 ° to 45 ° - cos = 1-32,The original type = 1? - 32? 2 = 43.10. Solution: 1 ° tan? Tan 2 °? Tan 3 °? Tan 4 °? ... ?Tan 87 °? Tan 88 °? Tan 89 ° = (? Tan tan 1 ° 89 °)? (tan 2 °, 88 °) tan? ... ? 44 ° (tan, tan, 46 °)? Tan 45 °= 1 * 1 * 1 *... * 1 * 1 = 1.Direct tests1. Solution: the original formula = 3 + 1-1 = 3.Solution: the original is equal to (3) -1 times 32-12 + 8 times 0.125So it's 13 times 32 minus 12 plus 1 is 1.28.2 solve right triangleClass 1Smart upgrade1. A2. A3. C4.125.456.1:27. Solution: (1) c = a2 + b2 = 42 + 82 = 45.(2) a = btan B = 103 = 1033,C = bsin B = 10 sin 60 ° = 1032 = 1032.A = c (3)? Sine of A is equal to 20 times 32 is 103,B = c? Cos 60 ° 20 x = 12 = 10.8. Solution: A and C respectively are AE orthogonal BC, CF perp AD, the feet are E and F, and AE = CF.In Rt delta ABE, AE is equal to AB? 45 ° sin = = 42 CF.In Rt delta CDE, CD = CFcos < FCD = 30 ° CFcos = 4232 = 863.9. Solution: (1) to prove: ∵ AD an BC,∴ delta ABD and delta ADC for a right triangle.∴ tan B = ADBD, cos < DAC = ADAC.∵ tan B = cos < DAC, ∴ ADBD = ADAC, namely the AC = BD.(2) in Rt delta ADC, we know that sin C = 1213,AD = 12k, AC = 13k.∴ CD = AC2 - AD2 = 5 k.∵ BC = BD + CD and AC = BD,∴ BC = 13 k + 5 k = 18 k.By the known BC = 12, ∴ 18 k = 12.∴ k = 23. ∴ AD = 12 k = 8.10. Solution: in the delta ADC, < = 90 °, ADC< = 60 ° C, AC = 10, CD = 5, AD = 53.In the delta of the ADB, < B = 45 °,< = 90 ° of ADB. BD = AD = 53,∴ AB = 5 6, BC = 5 (3 + 1).Direct tests1.62. M - n? Tan alpha tan alpha.3. Solution: M is MN perp AC, at this time MN is the smallest, AN =1 500 m.4. Solution: ∵ in Rt delta ADB, BDA, a < = 45 °, AB = 3,∴ DA = 3.CDA = ∵ in Rt delta ADC, < 60 °,∴ tan60 ° = CAAD. ∴ CA = 33.∴ BC = CA - BA = (33-3) m.A: the height of the traffic sign BC is (33-3) meters.The second classSmart upgrade1. B2. C3. C4. No5.306.11.8Solution: as shown in the picture, light FE affects E in the B building.EG: (1) EG: 1.< FEG = 30 °,The FG = 30 x 30 ° tan = 30 x = 103 = 3317.32 (m),MG = FM - GF = 20-17.32 = 2.68 (m).CD = ∵ DN = 2 m, 1.8 m,∴ ED = 2.68 2 = 0.68 (m).The shadow of A building affects the first floor of the B building, blocking the Windows of the building by 0.68 m.8. Solution: prolong BC to AD at E point, then CE perp AD.In Rt delta AEC, AC = 10, by the slope ratio of 1:3: < CAE = 30 °,∴ CE = AC? 30 ° sin = 10 x 12 = 5,AE = AC? Cos 30 ° 32 = = 10 x 53.In Rt delta ABE, BE = ab2-ae2= 142 -? 53? 2 = 11.∴ BC = BE - CE = 11-5 = 6 (m).A: the height of the flagpole is 6 meters.9. Solution:P for PC perp AB, perpendicular to C, then< = 30 °, APCAC = PC? Tan, 30 °,BC = PC? Tan 45 °.∴ PC, ∵ AC + BC = AB? Tan 30 ° + PC?Tan 45 ° = 100.∴ (33 + 1)? PC = 100.∴ PC = 50 x 50 (3-3) material (3-1.732)63.4 > 50 material.Answer: the distance between the center of the forest reserve and the line AB is greater than the radius of the reserve, so the highway planned to be built will not go through the protected area.Direct tests1. A.2. Solution: extend the CD to AB in G, and CG = 12 km,PC = 300 x 10 = 3 000 m = 3 km,In Rt delta PCD,CD = PC? Tan < P = 3 x tan60 ° = 33.∴ 12 - CD = 12-33 material 6.8 km.A: the mountain is about 6.8 kilometers high.According to the question, we can tell< < the ACB = 45 °, the ADB = 60 °, DC = 50.In Rt delta ABC,By < BAC = < BCA = 45 °, BC = AB.In Rt delta ABD, by tangent of ADB = ABBD,Have to BD = = ABtan < the ADB ABtan 60 ° = 33 ab.∵ BC - BD = DC again.∴ AB - 33 AB = 50, that is, (3-3) AB = 150.∴ AB = 1503-3 material 118.The perpendicular line of the BC is the perpendicular to the BC.By the question: < = 45 °, CAD < BAD = 60 °,AD = 60 m.In Rt delta ACD, < CAD = 45 °, AD an BC,∴ CD = AD = 60.In Rt delta ABD,∵ tan < BAD = BDAD,∴ BD = AD? Tan < BAD = 603.∴ BC = CD + BD= 60 + 603Material of 163.9 m.A: the tall building is about 163.9 m.The third classSmart upgrade1. D2.25033.90.6(1) building 1 height: AB = BD? Tan 30 ° 30 x =0.5774 material (m);High tower 2: CD + 30 x = 30 x 30 ° tan tan 5 ° material20 (m).(2). BD = CD present tan 30 ° 35 (m) = 20 present 0.5 774 material, so the distance between the two floor should be at least 35 m apart when will eliminate the impact.5. Solution: there is danger of hitting the rocks.Reason: past point P for PD perp AC at D.Let's say PD is x, in Rt delta PBD,< PBD = 45 °, 90 ° to 45 ° = ∴ BD = = x PD.In Rt delta PAD, ∵ < PAD = 30 °, 90 ° to 60 ° =∴ AD = 30 ° xtan = 3 x∴ ∵ AD = AB + BD, 3 x = 12 + x∴ x = 123-1 = 6 (3 + 1).6 (3 + 1) < 18 ∵∴ fishing boats not change course continue sailed east, there are dangerous on the rocks.6. Solution: according to the question, < A = 30 °, < PBC = 60 °.< APB. = 30 ° to 60 ° to 30 ° = ∴ < APB = < A∴ AB = PB.In Rt delta BCP, < = 90 ° C,< PBC = 60 °,PC = 450 m,∴ PB = 60 ° sine 450 = 450 = 3003 (m).∴ AB = PB = 3003 520 m material.Answer: (1) make a DG perp ABIn G, EH orthogonal complement AB to H. EHGD is rectangle∴ EH = DG = 5 m.∵ DGAG = 11.2, ∴ AG = 6 m.∵ EHFH = 11.4, ∴ FH = 7 m.∴ FA = = 7 FH + GH - AG + 1-6 = 2 (m)∴ S trapezoid ADEF = 12 + AF (ED)? EH = 12 (1 + 2) * 5 = 7.5 (m2),V = 7.5 x 4 000 = 30 000 (m3).∴ 30 000 m3 earthwork is needed to complete the project.(2) the party a originally planned to complete the x m3 soil each day, and party b had planned to complete the y m3 soil every day.According to the question, 20? X + y? Is equal to 3, 000, 15 [? 1 + 30%? X +? 1 + 40%? Y] = 30 000So if you simplify this, you get x plus y is equal to 1, 500, 1.3 x plus 1.4 y is equal to 2, 000.So the solution is x is equal to 1,000, y is equal to 500.A: the team originally planned to complete 1, 000 m3 of earth a day, and party b originally planned to complete 500 m3 of soil per day.Direct tests1. C2.11.23.90 °4. (40 + 403) nautical milesChapter 29 projection and view 29.1 projectionClass 1Smart upgrade1. B2. C3, C4. D5. B6.4.2 above8. Solution: connecting AE, point C as CF ∥ AE,For DF ∥ BE DF in point F, DF is the shadow of the CD, as shown in figure.The picture reflects the situation under the street lamp.(2) the light in the picture is parallel, and the light in the picture intersects at one point.(3) in the picture, the "AB" and "EF" are the lines of the little li in the sun and under the lamp.10. Solution: (1) figure.(2) by the question, to delta ABC ∽ delta GHC style, ∴ ABGH = BCHC.∴ 1.6 GH = 36 + 3. ∴ GH = 4.8 m.(3) delta A1B1C1 ∽ delta GHC1, ∴ A1B1GH = B1C1HC1.Set B1C1 to be x m, then 1.64.8 = xx + 3,So we solve for x is equal to 32 m, which is B1C1 is equal to 32 m.So it's the same thing as 1.64.8 is equal to B2C2B2C2 plus 2, so you have to solve for B2C2 is equal to 1 m.Direct tests1.3.32. C3. D4.11.8 mThe second classSmart upgrade1. C2. A3. B4. D5. The same6.127. The center is 8The quadrilateral ABCD is the orthographic projection of the cuboid.Section 10. Solution: ∵ cylindrical axis parallel to the plane, cylindrical orthogonal projection is side length is 4 square,∴ bottom, radius of the cylinder is 2, the high of 4.∴ the volume of a cylinder is PI * 22 * 4 = 16 PI,The surface area of the cylinder is 2 times PI times 22 plus 4 PI times 4 is 24 PI.11. Solution: as shown in the picture, C is used as the CE vertical flagpole AB at E, connecting AC.In Rt delta ACE, ∵ AECE = 11.5,∴ AE = 11.5 * 21= 14.∴ AB = AE + BE + 2 = = 14 16 (m).∴ flagpole is 16 meters high.Solution: as shown in the picture, AB is a bamboo pole, the CD is the shadow on the wall, lengthened the light and the AD is intersecting with the ground plane to E.The tangent of E is equal to the CDCE, which is 42 + CE = 1CE,So the solution is equal to 23.So xiao Ming should move the bamboo pole more than 23 meters.Direct testsD. 1. C2 and C3. B4.29.2 third viewSmart upgrade1. A2. D3. C4. A5. B6. A7. Solution: the number of cuboids in each position is marked in the top view, as shown. So there are 9 boxes.8. Solution: this geometric body is a cone, its bus length is1, and the base radius is 12.So the lateral area is PI times 1 times 12, which is 12 PI.9. Solution: (1) triangular prism. (2) as shown below.(3) the length of the main view is 10 cm, the side length of the triangle in the top view is 4 cm, and the height of the positive triprism is 10 cm, and the side length of the base side is 4 cm.So the side area is 4 times 3 times 10 is 120.Direct tests1. C2. A3. B4. B5. D6. A7. C8. C9. B10. B11.6。