如图,四边形EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则tan∠AHE的值为( ) A.15 B.310 C.16 D.27
问题描述:
如图,四边形EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则tan∠AHE的值为( )
A.
1 5
B.
3 10
C.
1 6
D.
2 7
答
∵四边形EFGH是矩形ABCD的内接矩形,EF:FG=3:1,AB:BC=2:1,
∴∠HEA+∠FEB=90°,
∵∠FEB+∠EFB=90°,
∴∠HEA=∠EFB,
∵∠HAE=∠B,
∴Rt△HAE∽△EBF,
∴
=HA EB
=AE FB
=HE EF
,1 3
同理可得,∠GHD=∠EFB,HG=EF,
∴△GDH≌△EBF,DH=BF,DG=EB,
设AB=2x,BC=x,AE=a,BF=3a,
则AH=x-3a,AE=a,
∴tan∠AHE=tan∠BEF,
即
=a x−3a
,解得:x=8a,3a 2x−a
∴tan∠AHE=
=a x−3a
=a 8a−3a
.1 5
故选A