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trigonometry
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http://show.simpload.com/082548b28714c5423.jpg?server=s3 1. find the angle between the planes AFH and EFGH. 2.Find the angle between the line BD and the plane BCFE. 3.Find the angle between the line AF and the plane BCFE. 4.Find the volume of tetrahedron ANBC. 5.Find the area of the shadow.
1.draw two diagonals on EFGH(畫個交叉呀) name the intersection point of two diagonals to be I the angle needed is angleAIE 不過你唔比FG幾長,計唔到 2.the angle needed is angleDBF AE/EB=tan30o EB=2/tan30o FB=square root(EF^2+EB^2) =square root(3^2+(2/tan30o)^2) =square root(9+12) =square root(21) tan angleDBF=DF/FB =2/(square root(21)) =0.4364 angleDBF=23.58o 3.the angle needed is angleAFH cos angleCBA=6/10 angleCBA=53.13o AH/AB= sin angleCBA =sin 53.13o =0.8 AH=0.8x6=4.8 DF=square root(EF^2-DE^2) =square root(10^2-6^2) =8 AF=square root(AD^2+DF^2) =square root(15^2+8^2) =17 finally,sin angleAFH=AH/AF =4.8/17 angleAFH=16.40o 3.(留意angleANC,angleCNB are right angles) NC=square root(AC^2-AN^2) =square root(7500)=86.60 what?你有冇畫錯?數據顯示angleNBC才是正角 重新開始:(留意angleANC,angleNBC are right angles) NC=square root(AC^2-AN^2) =square root(7500)=86.60 NB=square root(NC^2-BC^2) =square root(1100)=33.17 volume of tetrahedron=1/3 x base area x height =1/3(1/2NCxNB)x50 =1/3(1/2x86.60x33.17)x50 =23935.68cm^3 5.汗...一個距離也沒有...條問題又唔完整 AC/CD=tan30o CD=AC/tan30o using BC as the base, the height of triangleBCD=CD cos55o =ACcos55o/tan30o area of triangleBCD=1/2 BCxheight =1/2 BC x ACcos55o/tan30o
其他解答:
trigonometry
發問:
http://show.simpload.com/082548b28714c5423.jpg?server=s3 1. find the angle between the planes AFH and EFGH. 2.Find the angle between the line BD and the plane BCFE. 3.Find the angle between the line AF and the plane BCFE. 4.Find the volume of tetrahedron ANBC. 5.Find the area of the shadow.
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最佳解答:1.draw two diagonals on EFGH(畫個交叉呀) name the intersection point of two diagonals to be I the angle needed is angleAIE 不過你唔比FG幾長,計唔到 2.the angle needed is angleDBF AE/EB=tan30o EB=2/tan30o FB=square root(EF^2+EB^2) =square root(3^2+(2/tan30o)^2) =square root(9+12) =square root(21) tan angleDBF=DF/FB =2/(square root(21)) =0.4364 angleDBF=23.58o 3.the angle needed is angleAFH cos angleCBA=6/10 angleCBA=53.13o AH/AB= sin angleCBA =sin 53.13o =0.8 AH=0.8x6=4.8 DF=square root(EF^2-DE^2) =square root(10^2-6^2) =8 AF=square root(AD^2+DF^2) =square root(15^2+8^2) =17 finally,sin angleAFH=AH/AF =4.8/17 angleAFH=16.40o 3.(留意angleANC,angleCNB are right angles) NC=square root(AC^2-AN^2) =square root(7500)=86.60 what?你有冇畫錯?數據顯示angleNBC才是正角 重新開始:(留意angleANC,angleNBC are right angles) NC=square root(AC^2-AN^2) =square root(7500)=86.60 NB=square root(NC^2-BC^2) =square root(1100)=33.17 volume of tetrahedron=1/3 x base area x height =1/3(1/2NCxNB)x50 =1/3(1/2x86.60x33.17)x50 =23935.68cm^3 5.汗...一個距離也沒有...條問題又唔完整 AC/CD=tan30o CD=AC/tan30o using BC as the base, the height of triangleBCD=CD cos55o =ACcos55o/tan30o area of triangleBCD=1/2 BCxheight =1/2 BC x ACcos55o/tan30o
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