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如图,AB为⊙O的直径,弦CD⊥AB于E,∠CDB=15°,OE=23.(1)求⊙O的半径;(2)将△OBD绕O点旋转,使弦
如图,AB为⊙O的直径,弦CD⊥AB于E,∠CDB=15°,OE=23.(1)求⊙O的半径;(2)将△OBD绕O点旋转,使弦BD的一个端点与弦AC的一个端点重合,则弦BD...
如图,AB为⊙O的直径,弦CD⊥AB于E,∠CDB=15°,OE=23.(1)求⊙O的半径;(2)将△OBD绕O点旋转,使弦BD的一个端点与弦AC的一个端点重合,则弦BD与弦AC的夹角为______.
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(1)∵AB为⊙O的直径,弦CD⊥AB于E,
∴弧BC=弧BD,
∴∠BDC=
∠BOD,
而∠CDB=15°,
∴∠BOD=2×15°=30°,
在Rt△ODE中,∠DOE=30°,OE=2
,
∴OE=
DE,OD=2DE,
∴DE=
=2,
∴OD=4,
即⊙O的半径为4;
(2)有4种情况:如图:
![](https://iknow-pic.cdn.bcebos.com/0e2442a7d933c8953a8b51cad21373f08302008e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://iknow-pic.cdn.bcebos.com/83025aafa40f4bfb082557e5004f78f0f636188e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://iknow-pic.cdn.bcebos.com/aa64034f78f0f736570b85b00955b319eac4138e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://iknow-pic.cdn.bcebos.com/ca1349540923dd54f5fe6ed9d209b3de9d8248f6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
①如图1所示:∵OA=OB,∠AOB=30°,
∴∠OAB=∠OBA=75°,
∵CD⊥AB,AB是直径,
∴弧BC=弧BD,
∴∠CAB=
∠BOD=15°,
∴∠CAB=∠BAO+∠CAB=15°+75°=90°;
②如图2所示,∠CAD=75°-15°=60°;
③如图3所示:∠ACB=90°;
④如图4所示:∠ACB=60°;
故答案为:60°或90°.
∴弧BC=弧BD,
∴∠BDC=
1 |
2 |
而∠CDB=15°,
∴∠BOD=2×15°=30°,
在Rt△ODE中,∠DOE=30°,OE=2
3 |
∴OE=
3 |
∴DE=
2
| ||
|
∴OD=4,
即⊙O的半径为4;
(2)有4种情况:如图:
①如图1所示:∵OA=OB,∠AOB=30°,
∴∠OAB=∠OBA=75°,
∵CD⊥AB,AB是直径,
∴弧BC=弧BD,
∴∠CAB=
1 |
2 |
∴∠CAB=∠BAO+∠CAB=15°+75°=90°;
②如图2所示,∠CAD=75°-15°=60°;
③如图3所示:∠ACB=90°;
④如图4所示:∠ACB=60°;
故答案为:60°或90°.
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