Question

愛玉

求解

205
wiz
12 FF
5. 已知兩點 A(2cosa,2sina),B(cosp,sin),且a-B=60°,則線段AB的長為
2005a
Sina
M =
cosp
Sinß
$
T
372
COSB
cosa
tan 60°: 3
sin a
=
sinp
cosa
COSB
sind sing
tand
tanẞ
2
1 +
tand tanp
^
=(+)200=nia-tan (90-α) - tanc (

Answer 1

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Davis

AB^=(cosB-2cos Alpha)^2+(sin B-2sin Aipha)^2=cosB^2-4cosAlpha cosB+4cos Alpha^2+sinB^2-4sin Alpha sinB+4sin Alpha^2=5-4cos(Alpha-B)
=5-4cos 60=5-2=3

So AB=根號3