∵OB=OQ,
∴∠OBP=∠OQB,
∵OA⊥OB,
∴∠BQA=
1 |
2 |
1 |
2 |
∵EQ是切线,
∴∠OQE=90°,
∴∠OBP+∠AQE=∠OQB+∠AQE=90°-∠BQA=90°-45°=45°;
(2)如图②,连接OQ,
∵OB=OQ,
∴∠OBQ=∠OQB,
∵OA⊥OB,
∴∠BQA=
1 |
2 |
∴∠OQA=∠BQA-∠OQB=135°-∠OBQ,
∵EQ是切线,
∴∠OQE=90°,
∴135°-∠OBQ+∠AQE=90°,
整理得,∠OBQ-∠AQE=45°,
即∠OBP-∠AQE=45°.