如图1，点O为直线AB上一点，过点O作射线OC，使∠AOC=60°．将一把直角三角尺的直角顶点放在点O处，一边OM在射线OB上，另一边ON在直线AB的下方，其中∠OMN=30°．（1）将图1中的三角尺绕点O顺时针旋转至图2，使一边OM在∠BOC的内部，且恰好平分∠BOC，求∠CON的度数；（2）将图1中的三角尺绕点O按每秒10°的速度沿顺时针方向旋转一周，在旋转的过程中，在第______秒时，边MN恰好与射线OC平行；在第______秒时，直线ON恰好平分锐角∠AOC．（直接写出结果）；（3）将图1中的三角尺绕点O顺时针旋转至图3，使ON在∠AOC的内部，请探究∠AOM与∠NOC之间的数量关系，并说明理由．扫码下载作业帮搜索答疑一搜即得答案解析 查看更多优质解析（1）∵∠AOC=60°，∴∠BOC=120°，又∵OM平分∠BOC，∴∠COM=∠BOC=60°，∴∠CON=∠COM+90°=150°；（2）∵∠OMN=30°，∴∠N=90°-30°=60°，∵∠AOC=60°，∴当ON在直线AB上时，MN∥OC，旋转角为90°或270°，∵每秒顺时针旋转10°，∴时间为9或27，直线ON恰好平分锐角∠AOC时，旋转角为90°+30°=120°或270°+30°=300°，∵每秒顺时针旋转10°，∴时间为12或30；故答案为：9或27；12或30．（3）∵∠MON=90°，∠AOC=60°，∴∠AON=90°-∠AOM，∠AON=60°-∠NOC，∴90°-∠AOM=60°-∠NOC，∴∠AOM-∠NOC=30°，故∠AOM与∠NOC之间的数量关系为：∠AOM-∠NOC=30°．（1）根据邻补角的定义求出∠BOC=120°，再根据角平分线的定义求出∠COM，然后根据∠CON=∠COM+90°解答；（2）分别分两种情况根据平行线的性质和旋转的性质求出旋转角，然后除以旋转速度即可得解；（3）用∠AOM和∠CON表示出∠AON，然后列出方程整理即可得解．本题考点：旋转的性质．考点点评：本题考查了旋转的性质，角平分线的定义，平行线的性质，读懂题目信息并熟练掌握各性质是解题的关键，难点在于（2）要分情况讨论．解析看不懂？免费查看同类题视频解析查看解答}

如图1，O为直线AB上一点，过点O作射线OC，∠AOC=30°，将一直角三角板（∠D=30°）的直角顶点放在点O处，一边OE在射线OA上，另一边OD与OC都在直线AB的上方．（1）将图1中的三角板绕点O以每秒5°的速度沿顺时针方向旋转一周，如图2，经过t秒后，OD恰好平分∠BOC．①此时t的值为___；（直接填空）②此时OE是否平分∠AOC？请说明理由；（2）在（1）问的基础上，若三角板在转动的同时，射线OC也绕O点以每秒8°的速度沿顺时针方向旋转一周，如图3，那么经过多长时间OC平分∠DOE？请说明理由；（3）在（2）问的基础上，经过多长时间OC平分∠DOB？请画图并说明理由．扫码下载作业帮搜索答疑一搜即得答案解析 查看更多优质解析（1）①∵∠AOC=30°，∠AOB=180°，∴∠BOC=∠AOB-∠AOC=150°，∵OD平分∠BOC，∴∠BOD=12∠BOC=75°，∴t=90°-75°5=3．②是，理由如下：∵转动3秒，∴∠AOE=15°，∴∠COE=∠AOC-∠AOE=15°，∴∠COE=∠AOE，即...解析看不懂？免费查看同类题视频解析查看解答}

已知∠AOD＝40°，射线OC从OD出发，绕点O以20°/秒的速度逆时针旋转，旋转时间为t秒.射线OE、OF分别平分∠AOC、∠AOD.
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综合题
（1）
如图①：如果t＝4秒，求∠EOA的度数；
（2）
如图①：若射线OC旋转时间为t（t≤7）秒，求∠EOF的度数（用含t的代数式表示）；
（3）
若射线OC从OD出发时，射线OB也同时从OA出发，绕点O以60°/秒的速度逆时针旋转，射线OC、OB在旋转过程中（t≤3），
请你借助图②与备用图进行分析后，
（i）求此时t的值；（ii）
求的值.
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