已知一三角形ABC,〈A=20度,AB=AC,点E为AB上一点,且AE=BC,求〈BEC的度数。
来源:百度知道 编辑:UC知道 时间:2024/05/16 23:57:13
解答:
连接CE,在CE上做CG=AE,则BC=CG
三角形ABC内:〈A=20,则〈B=〈C=80
三角形BCG内:〈CGB=〈CBG=50
则:〈EBG=〈B-〈CBG=30
又因为〈BEC+〈EBG=〈CGB
故:〈BEC=〈CGB-〈EBG=50-30=20
∠BEC=30°
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∵AB=AC,∠A=20°
∴∠ABC=∠ACB=80°
设∠ACE=α,∠BEC=β,则有:
β=α+20°,α=β-20°,∠AEC=180°-β
在△ACE中,由正弦定理,得:
AE/sin(α)=AC/sin(180°-β)=AC/sin(β)…………(1)
同理,在△ABC中,有:
BC/sin(20°)=AC/sin(80°)…………………………(2)
已知AE=BC,(2)/(1),得:
sin(β-20°)/sin(20°)=sin(β)/sin(80°)………(3)
∵sin(80°)=cos(10°)
sin(20°)=2·sin(10°)·cos(10°)
sin(β-20°)=sin(β)·cos(20°)-cos(β)·sin(20°)
∴(3)式化为:
[sin(β)·cos(20°)-cos(β)·sin(20°)]/sin(β)=2·sin(10°)·cos(10°)/cos(10°)
即:cos(20°)-cos(β)·sin(20°)=2·sin(10°)
∴cot(β)=[cos(20°)-2·sin(10°)]/sin(20°)=cot(20°)-1/cos(10°)
∴β=arctan{1/[cot(20°)-1/cos(10°)]}=30°