题目内容

已知锐角△ABC内有一点O,满足OA=OB=OC,且∠A=60°,若
cosB
sinC
AB
+
cosC
sinB
AC
=2m
AO
,则m等于(  )
分析:把已知式两边同时乘以
OA
,设 OA=OB=OC=R,由于∠AOB=2C,∠AOC=2B,可得
cosB
sinC
 2(cos2C-1)
+
cosC
sinB
  2 (cos2B-1)
=-2mR2,由 m=sinBcosC+sinCcosB=sin(B+C)=sinA 求得结果.
解答:解:∵
cosB
sinC
AB
+
cosC
sinB
AC
=2m
AO

cosB
sinC
(
OB
-
OA
)+
cosC
sinB
(
OC
-
OA
)
=2m
AO

两边同时乘以
OA
 可得
cosB
sinC
(
OB
-
OA
)•
OA
+
cosC
sinB
(
OC
-
OA
)
OA
=2m
AO
OA

设 OA=OB=OC=R,由于∠AOB=2C,∠AOC=2B,
cosB
sinC
 2(cos2C-1)
+
cosC
sinB
  2 (cos2B-1)
=-2mR2
∴m=sinBcosC+sinCcosB=sin(B+C)=sinA=
3
2

故选B.
点评:本题主要考查两个向量的加减法的法则,以及其几何意义,诱导公式的应用,属于中档题.
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