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62 lines
2.7 KiB
62 lines
2.7 KiB
2 years ago
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#### (1) 用已知换未知就是办法
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$\angle AOD = \angle AOB+\angle BOD$
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$=60^{\circ}+\angle BOD$
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$\angle BOC=\angle COD-\angle BOD$
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=$90^{\circ}-\angle BOD$
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$\angle AOD+ \angle BOC=60^{\circ}+90^{\circ}+\angle BOD-\angle BOD=150^{\circ}$
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$\angle BOD-\angle AOC=\angle COD-\angle BOC-(\angle AOB- \angle BOC)=90^{\circ}-60^{\circ}=30^{\circ}$
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#### (2)
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$\angle MON=x+y+\angle BOC=\frac{1}{2}(\angle AOC+\angle BOD)+\angle BOC$
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$=\frac{1}{2}(\angle AOB -\angle BOC+\angle COD-\angle BOC)+\angle BOC$
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$=\frac{1}{2}(90^{\circ}+60^{\circ})=75^{\circ}$
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#### (3)常见的第二种问题,一般会有一个未知数
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提出要求,求解未知数,
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**解决方法:**
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1. 找出目标角
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2. 表示目标角
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3. 根据要求列方程求解
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- 任意时刻旋转的角度都是$3t$
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- $\angle AOB=60^{\circ}$,到$OD$止,$OB$共需走的角度是$120^{\circ}$,也就是,当$t=40$时,走完全程。
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- 重点是表示$\angle BOC$和$\angle MON$,这是 <font color='red' size=4><b> 重难点</b></font>
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$\because \angle AOB=60^{\circ},\angle AOC=90^{\circ}$
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$\therefore \angle BOC=90^{\circ}-60^{\circ}=30^{\circ}$
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当$OB$开始旋转时,分两种情况:
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- $OB'$在$OC$右侧,此时$\angle B'OC=30^{\circ}-3t$
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- $OB'$在$OC$左侧,此时$\angle B'OC=3t-30^{\circ}$
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- 算一下临界值:$30^{\circ}-3t=0^{\circ}$
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此时$t=10$,表示$10$秒之内,在右侧,等于$10$秒时,与$OC$重合
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- 第二种情况是在$10\sim 40$秒之间,此时 <font color='red' size=4><b>完全符合第二问的答案,即 $OC$在 $\angle AOB$内部!此时,$\angle MON=75^{\circ}$</b></font>
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情况1中$\angle MON$是多大呢?借鉴第二问的思路:
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$\angle MON=x+y-\angle BOC=\frac{1}{2}(\angle AOC+\angle BOD)-\angle BOC$
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$=\frac{1}{2}(\angle AOB +\angle BOC+\angle COD+\angle BOC)-\angle BOC$
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$=\frac{1}{2}(90^{\circ}+60^{\circ})=75^{\circ}$
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即:两种情况下,都是$\angle MON=75^{\circ}$
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$5(30-3t)=75^{\circ}$
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或
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$5(3t-30)=75^{\circ}$
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解得$t=5$或$t=15$
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依题意,$\angle AOB$在一同旋转,$OB$走多少度,$OA$就走多少度,而$OD$是不动的,所以$180^{\circ}$减去走的角度即可:
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$\angle AOD=180^{\circ}-3*5=165^{\circ}$
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或
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$\angle AOD=180^{\circ}-3*15=135^{\circ}$
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