✔ 最佳答案
1)滑梯高度= 6sin30=3
滑梯腳與滑梯底的距離=6cos30=5.196
單車與滑梯底的距離=3/tan12=14.11
so, 單車與滑梯腳的距離=14.114 -5.196 =8.92m
(你條問題同圖...唔相符既,,希望你明啦...)
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2)(a)角BAC =35+30=75
(b)由A畫下條垂直線落黎,,同BC線既交點係D...
角ABC=90-35=55
BD=60cos55=34.415m
角ACB=90-30=60
BD=75cos60=37.5m
B與C兩點的距離=37.5+34.4=71.9m
OR你想快D就直接寫...
B與C兩點的距離=75cos60+60cos55=71.9m
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3)遠d果個男孩係A,,近d係B,,"他"係C,,樹底係D,,
tan36=CD/BD
tan30=CD/AD=CD/BD+4
解左2條式...(我唔想打...)
應該CD=tan30/(1- tan30/tan36)=2.817m
or你可以搵左BD出黎先既...
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4)由A畫條線去AB線,,交點為C...
角CPA=155-90=65
AC=65sin65=58.91m
B與A兩點的距離=3*60=180m
BC=180-58.91=121.09m
PC=65cos65=27.47m
tan角BPC=121.09/27.47
角BPC=77.22
so,角BPA =77.22+65=142.22
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5)由A畫條垂直線落黎,,同BP線既交點係C...
角CAP=55
AC=100cos55=57.358
CP=100sin55=81.915m
BC=120-81.915=38.085
AB^2=BC^2+AC^2
AB=68.85m
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6)a)由P畫條橫線去左邊,,同AB伸延線既交點係C...
角CPB=90-50=40
角CPA=90-35=55
角BPA=55-40=15
b)tan35=PC/AC=PC/(1200+BC)
tan50=PC/BC
解方程,,得出BC=1709.4m
AP=(1709.4+1200)/cos30=3359m
(究竟係50定59,,我用左50)