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七年级数学解答题特难
题目

如图,点CCDD是半圆弧上的两个动点,在运动的过程中保持COD=90∠COD=90^{\circ}

(1)(1)如图OEOE平分AOC∠AOCOFOF平分BOD∠BOD,求EOF∠EOF的度数;

(2)(2)如图OEOE平分AOD∠AODOFOF平分BOC∠BOC,求EOF∠EOF的度数.

知识点:圆锥的认识,角的计算章节:未标注

答案与解析

答案

解:(1)OE(1)∵OE平分AOC∠AOCOFOF平分BOD∠BOD

COE=12AOC∴∠COE= \dfrac{1}{2}∠AOCDOF=12BOD∠DOF= \dfrac{1}{2}∠BOD

COD=90∵∠COD=90^{\circ}

COA+DOB=180COD=90∴∠COA+∠DOB=180^{\circ}-∠COD=90^{\circ}

COE+DOE=12(COA+DOB)=45∴∠COE+∠DOE= \dfrac{1}{2}\left(∠COA+∠DOB\right)=45^{\circ}

EOF=COE+DOF+COD=135∴∠EOF=∠COE+∠DOF+∠COD=135^{\circ}

(2)OE(2)∵OE平分AOD∠AODOFOF平分BOC∠BOC

AOE=12AOD∴∠AOE= \dfrac{1}{2}∠AODBOF=12BOC∠BOF= \dfrac{1}{2}∠BOC

EOF=AOFAOE∴∠EOF=∠AOF-∠AOE

=180BOF12AOD=180^{\circ}-∠BOF- \dfrac{1}{2}∠AOD

=18012BOC12AOD=180^{\circ}- \dfrac{1}{2}∠BOC- \dfrac{1}{2}∠AOD

=18012(BOC+AOD)=180^{\circ}- \dfrac{1}{2}\left(∠BOC+∠AOD\right)

=18012×(180COD)=180^{\circ}- \dfrac{1}{2}×\left(180^{\circ}-∠COD\right)

=9012COD=90^{\circ}- \dfrac{1}{2}∠COD

=45=45^{\circ}

解析

解:(1)OE(1)∵OE平分AOC∠AOCOFOF平分BOD∠BOD

COE=12AOC∴∠COE= \dfrac{1}{2}∠AOCDOF=12BOD∠DOF= \dfrac{1}{2}∠BOD

COD=90∵∠COD=90^{\circ}

COA+DOB=180COD=90∴∠COA+∠DOB=180^{\circ}-∠COD=90^{\circ}

COE+DOE=12(COA+DOB)=45∴∠COE+∠DOE= \dfrac{1}{2}\left(∠COA+∠DOB\right)=45^{\circ}

EOF=COE+DOF+COD=135∴∠EOF=∠COE+∠DOF+∠COD=135^{\circ}

(2)OE(2)∵OE平分AOD∠AODOFOF平分BOC∠BOC

AOE=12AOD∴∠AOE= \dfrac{1}{2}∠AODBOF=12BOC∠BOF= \dfrac{1}{2}∠BOC

EOF=AOFAOE∴∠EOF=∠AOF-∠AOE

=180BOF12AOD=180^{\circ}-∠BOF- \dfrac{1}{2}∠AOD

=18012BOC12AOD=180^{\circ}- \dfrac{1}{2}∠BOC- \dfrac{1}{2}∠AOD

=18012(BOC+AOD)=180^{\circ}- \dfrac{1}{2}\left(∠BOC+∠AOD\right)

=18012×(180COD)=180^{\circ}- \dfrac{1}{2}×\left(180^{\circ}-∠COD\right)

=9012COD=90^{\circ}- \dfrac{1}{2}∠COD

=45=45^{\circ}

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