题目内容
P是⊙O的直径AB的延长线上一点,PC与⊙O相切于点C,∠APC的平分线交AC于Q,则∠PQC=______.
连接OC,
∵PC与⊙O相切于点C,
∴OC⊥PC,
∵OA=OC,
∴∠OAC=∠OCA=
∠POC,
又∵∠APQ=∠CPQ=
∠APC,
PAC+∠APQ,
=
(∠POC+∠APC),
=
×90°,
=45°.
故答案为45°.
∵PC与⊙O相切于点C,
∴OC⊥PC,
∵OA=OC,
∴∠OAC=∠OCA=
1 |
2 |
又∵∠APQ=∠CPQ=
1 |
2 |
PAC+∠APQ,
=
1 |
2 |
=
1 |
2 |
=45°.
故答案为45°.
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