图①
C
C
A
图②
C E A
设AB=5a,BC=3a,则AC=4a
如图,在AB上取AD=AC=4a,作DE⊥AC于点E。
则DE=AD·sinA=4a·=CE=4a-
312416a,AE= AD·cosA=4a·=a 5555164a=a
5522224?4??12?CD?CE?DE??a???a??10 5?5??5?∴sadA?CD10 ?AC55. (2011广东东莞,19,7分)如图,直角梯形纸片ABCD中,AD∥BC,∠A=90°,∠C=30°.折叠纸片使BC经过点D.点C落在点E处,BF是折痕,且BF= CF =8. (l)求∠BDF的度数; (2)求AB的长.
【解】(1)∵BF=CF,∠C=30,
∴∠FBC=30,∠BFC=120 又由折叠可知∠DBF=30 ∴∠BDF=90 (2)在Rt△BDF中, ∵∠DBF=30,BF=8 ∴BD=43 ∵AD∥BC,∠A=90 ∴∠ABC=90
00000000
又∵∠FBC=∠DBF=30 ∴∠ABD=30 在Rt△BDA中,
∵∠AVD=30,BD=43 ∴AB=6.
6. (2011湖北襄阳,19,6分)
1x2?2x?1先化简再求值:(,其中x?tan60??1. ?1)?x?2x2?4000【答案】 原式??x?1(x?2)(x?2)x?2??? ········································································· 2分 2x?2(x?1)x?1 当x?tan60??1?3?1时, ····················································································· 3分
原式??
3?1?23?1?13?33?3?1. 6分
??