Question

已知△ABC的外接圆半径为1，圆心为O，且3

OA

+4

OB

+5

OC

=0，则∠AOB=（　　）

• A、

  π

  6

• B、

  π

  4

• C、

  π

  3

• D、

  π

  2

Answer 1

分析：由△ABC的外接圆半径为1，圆心为O，可得|

OA

|=|

OB

|=|

OC

|=1．由于3

OA

+4

OB

+5

OC

=0，可得3

OA

+4

OB

=-5

OC

．两边作数量积得(3

OA

+4

OB

)2=(-5

OC

)2，整理即可得出．

解答：解：∵△ABC的外接圆半径为1，圆心为O，∴|

OA

|=|

OB

|=|

OC

|=1．
∵3

OA

+4

OB

+5

OC

=0，∴3

OA

+4

OB

=-5

OC

．
两边作数量积得(3

OA

+4

OB

)2=(-5

OC

)2，
∴9

OA

2+16

OB

2+24

OA

•

OB

=25

OC

2，
∴24

OA

•

OB

=0，
∴

OA

⊥

OB

．
∴∠AOB=

π

2

．
故选：D．

点评：本题考查了三角形外接圆的性质、数量积运算、向量垂直与数量积的关系等基础知识与基本技能方法，属于难题．