MATHS(三角學2) @ dlxpxv7的部落格

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MATHS(三角學2)

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7.圖中,角ABC= 30度 ,角ADC= 45度,且AB=AC=開方2 求CDhttp://s731.photobucket.com/albums/ww315/cc2lamlam1314/?action=view¤t=DSC00093.jpg8. 在三角形ABC中,設a,b,c分別為A,B ,C 對邊的長度。若 sin A: sinB: sin C = 4:5:6則A. A:B:C = 4:5:6B. a:b:c=4:5:6 C. tanA:tanB:tanC=4:5:6D.... 顯示更多 7.圖中,角ABC= 30度 ,角ADC= 45度,且AB=AC=開方2 求CD http://s731.photobucket.com/albums/ww315/cc2lamlam1314/?action=view¤t=DSC00093.jpg 8. 在三角形ABC中,設a,b,c分別為A,B ,C 對邊的長度。若 sin A: sinB: sin C = 4:5:6則 A. A:B:C = 4:5:6 B. a:b:c=4:5:6 C. tanA:tanB:tanC=4:5:6 D. (b+c):(c+a):(a+b)=4:5:6 9.圖中所示為一個正方體,且C為MN的中點。求cos角ACB http://s731.photobucket.com/albums/ww315/cc2lamlam1314/?action=view¤t=DSC00054.jpg 10. 圖中,AB為一枝直立的旗桿,P和Q分別在旗桿的北面和東面。若AB = 1 , BP = 2 ,BQ=3 求cos角PAQ http://s731.photobucket.com/albums/ww315/cc2lamlam1314/?action=view¤t=DSC00055.jpg

最佳解答:

7. ABC is an isosceles triangle(等腰三角形), angleABC = angleACB = x. Sum of interior angles of triangle = 30 + x + x = 180 => x = 75 Exterior angle of triangle ACB = CAD + CDA 75 = CAD + 45 CAD = 30 By sine rule, AC / sinCDA = CD / sinCAD √2 / sin45 = CD / sin30 √2/(√2/2)*(1/2) = CD CD = 1 8. (B) a:b:c = 4:5:6 9. Let side of the cube be L. CN = L/2; BN = L CB2 = CN2 + BN2 CB2 = L2/4 + L2 CB = (√5/2)L CM = L/2; AM = L AC2 = CM2 + AM2 AC = (√5/2)L AB2 = AN2 + BN2 AB2 = (AM2 + MN2) + BN2 AB2 = L2 + L2 + L2 AB = √3L Cosine rule, AB2 = AC2 + CB2 – 2(AC)(BC)cosACB 3L2 = 5/4L2 + 5/4L2 – 2(5/4)L2(cosACB) cos ACB = [(5/4) + (5/4) – 3] / (5/2) cos ACB = -0.2 ACB = 101.5 degree 10. AP2 = AB2 + BP2 AP2 = 1 + 4 AP = √5 AQ2 = AB2 + BQ2 AQ2 = 1 + 9 AQ = √10 QP2 = QB2 + BP2 QP2 = 9 + 4 QP = √13 cosine rule: cosPAQ = (AQ2 + AP2 – QP2)/[2(AQ)(AP)] cos PAQ = (10 + 5 – 13)/(2)(√10)√5) = 0.1414 PAQ = 81.9

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