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1. 「★☆」如图,在$△ ABC$中,$∠ BAC=120°$,将$△ ABC$绕点A按逆时针方向旋转得到$△ AB'C'$,若点$B'$恰好落在$BC$边上,且$AB'=CB'$,则$∠ C'$的度数为 (
A.$18°$
B.$20°$
C.$22°$
D.$24°$
B
)A.$18°$
B.$20°$
C.$22°$
D.$24°$
答案
1.B 由旋转的性质可知 $AB=AB',∠C=∠C',\therefore ∠B=∠AB'B$,
$\because AB' = CB',\therefore ∠C = ∠CAB',\therefore ∠B = ∠AB'B = ∠C + ∠CAB' = 2∠C,\because ∠BAC+∠C+∠B = 180°,∠BAC = 120°,$
$\therefore ∠C+∠B = 3∠C = 60°,\therefore ∠C = 20°,\therefore ∠C' = 20°$. 故选 B.
$\because AB' = CB',\therefore ∠C = ∠CAB',\therefore ∠B = ∠AB'B = ∠C + ∠CAB' = 2∠C,\because ∠BAC+∠C+∠B = 180°,∠BAC = 120°,$
$\therefore ∠C+∠B = 3∠C = 60°,\therefore ∠C = 20°,\therefore ∠C' = 20°$. 故选 B.
2.「2026广东广州花都期中，★★☆」如图，把$\mathrm{Rt}△ ABC$绕点A逆时针旋转$40°$，得到$\mathrm{Rt}△ AB'C'$，点$C'$恰好落在边AB上，连接$BB'$，则$∠ BB'C'=$
20
度.答案
2.答案 20
解析 由旋转的性质得 $AB = AB',∠AC'B' = ∠C = 90°,$
$\therefore ∠ABB' = ∠AB'B = \frac{1}{2}(180°-∠BAB') = \frac{1}{2}×(180°-40°) = 70°,\therefore ∠BB'C' = 90°-∠ABB' = 90°-70° = 20°.$
解析 由旋转的性质得 $AB = AB',∠AC'B' = ∠C = 90°,$
$\therefore ∠ABB' = ∠AB'B = \frac{1}{2}(180°-∠BAB') = \frac{1}{2}×(180°-40°) = 70°,\therefore ∠BB'C' = 90°-∠ABB' = 90°-70° = 20°.$
3.「2026浙江杭州西湖期中，★★☆」如图，在$△ ABC$中，$∠ CAB = 20°$，$∠ ABC = 30°$，将$△ ABC$绕A点逆时针旋转$50°$得到$△ AB'C'$，则$∠ BB'C' =$
$95°$
。答案
3.答案 $95°$
解析 由旋转的性质得 $∠BAB' = 50°,AB = AB',∠AB'C' = ∠ABC = 30°,\therefore ∠AB'B = ∠ABB' = \frac{1}{2}×(180°-50°) = 65°,$
$\therefore ∠BB'C' = ∠AB'B+∠AB'C' = 65°+30° = 95°.$
解析 由旋转的性质得 $∠BAB' = 50°,AB = AB',∠AB'C' = ∠ABC = 30°,\therefore ∠AB'B = ∠ABB' = \frac{1}{2}×(180°-50°) = 65°,$
$\therefore ∠BB'C' = ∠AB'B+∠AB'C' = 65°+30° = 95°.$
4.「2026 海南海口期末，★★☆」如图,在$△ ABC$中,$∠ ACB = 90°$,将$△ ABC$绕点A顺时针旋转得到$△ ADE$,点B,C的对应点分别为点D,E,DE的延长线与边BC相交于点F,连接CE.若$AC = 2$,$CF = 1$,则线段CE的长为 (
A.$\frac{2\sqrt{5}}{5}$
B.$\frac{3\sqrt{5}}{5}$
C.$\frac{4\sqrt{5}}{5}$
D.$\sqrt{5}$
C
)A.$\frac{2\sqrt{5}}{5}$
B.$\frac{3\sqrt{5}}{5}$
C.$\frac{4\sqrt{5}}{5}$
D.$\sqrt{5}$
答案
4.C 如图,连接 AF 交 CE 于点 G,
根据旋转的性质可得 $AE = AC = 2$,
$∠AED = ∠ACF = 90°,$
$\therefore ∠AEF = 180°-90° = 90°,$
由勾股定理得 $AF = \sqrt{AC^2+CF^2} = \sqrt{5},$
$\therefore EF = \sqrt{AF^2-AE^2} = 1,$
$\therefore CF = EF.\because AE = AC,\therefore AF$ 垂直平分 $CE,\therefore CG = EG.$
$\because S_{△ ACF} = \frac{1}{2}AC · CF = \frac{1}{2}AF · CG,$
$\therefore CG = \frac{AC · CF}{AF} = \frac{2\sqrt{5}}{5},\therefore CE = 2CG = \frac{4\sqrt{5}}{5}$.故选 C.
5.「2026 山东枣庄台儿庄期中，★★☆」如图，在正方形ABCD中，将边BC绕点B逆时针旋转得到BE，∠CED=90°, ED=1, 则线段BE的长度为
$\sqrt{5}$
。答案
5.答案 $\sqrt{5}$
解析 如图,过点 B 作 $BF ⊥ CE$,垂足为 $F,\therefore ∠BFC = 90°,$
$\therefore ∠BCF+∠FBC = 90°,\because$ 四边形 ABCD 为正方形,$\therefore BC = CD,∠BCD = 90°,\therefore ∠BCF+∠DCE = 90°,\therefore ∠DCE = ∠FBC.$
在 $△ BCF$ 和 $△ CDE$ 中, $\begin{cases} ∠BFC = ∠CED = 90°, \\ ∠CBF = ∠DCE, \\ BC = CD, \end{cases}$ $\therefore △ BCF ≌ △ CDE(AAS),\therefore CE = BF,CF = DE = 1$,由旋转的性质可知 $BE = BC,\because BF ⊥ CE,\therefore EF = CF = 1,\therefore BF = CE = EF+CF = 2,$
$\therefore BE = \sqrt{BF^2+EF^2} = \sqrt{5}.$
