7.在圖中，AB=BD=DF=7cm，FG=24cm及BC//DE//FG
a.求DP和DE的長度
b.由此，求BC的長度
圖：(&b=2&(這個位係冇;的)f=190113)
http://www.wretch.cc/album/show.php?i=Puchikoyiu&b=2&f=1901133632&p=5
（係下面果張）↑
８．sin2次+sin2次18度
９．tan15度xtan75度+sin15度/cos75度
請各大大幫助求求小的,感激不盡!
數學大難題
2007-07-19 12:37 am
回答 (3)
2007-07-19 5:08 pm
✔ 最佳答案
7a)Consider triangle BFG.BP/PG = BD/DF = 1 (intercept theorem)
BP=PG
So, DP = 1/2FG = 12 cm,Ans.(mid-pt. theorem)
b)Consider triangle BCG.
BP/PG = CE/EG = BD/DF = 1 (intercept theorem)
BP=PG and CE=EG
So, PE=1/2 BC
Consider triangle ADE.
DE = 2BC (mid-pt. theorem)
DP + PE = 2BC
12 + 1/2BC = 2BC
BC = 12/ (3/2) = 8 cm,Ans.
但係要先找DE的長度來計我就唔識喇...
8) sin^2 72 + sin^2 18
= sin^2 (90 - 18) + sin^2 18
= cos^2 18 + sin^2 18
=1,Ans.
9) tan15 x tan75 +sin15/cos75
=tan15 x tan(90 - 15) + sin15/cos(90 - 15)
= tan15/tan15 + sin15/sin15
=2,Ans.
2007-07-19 12:55 am
7.無法看圖
8.sin2次(呢度係咪仲有ga?)+sin2次18度,未見過甘ge題.
9. tan15度xtan75度+sin15度/cos75度
=tan15x(1/tan15)+sin15/sin15
=1+1
=2 (so easy!)
8.sin2次(呢度係咪仲有ga?)+sin2次18度,未見過甘ge題.
9. tan15度xtan75度+sin15度/cos75度
=tan15x(1/tan15)+sin15/sin15
=1+1
=2 (so easy!)
2007-07-19 12:41 am
呢個....你果個網都開唔到....冇圖唔識計...= =
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