[Toán 11] Tìm góc BMC

V

vy000

Dựng về phía ngài $\Delta ABC$ một $\Delta ADB$ sao cho $\Delta ADB=\Delta AMC$

$\Rightarrow \begin{cases} \widehat{DAB}=\widehat{MAC}\Rightarrow \widehat{DAM}=\widehat{BAC}=60^o\\AD=
AM\\DB=CM \Rightarrow DB^2+BM^2=MA^2\end{cases}$

$\Rightarrow \begin{cases} \Delta ADM \text{đều} \Rightarrow MD=MA\\ DB^2+BM^2=MA^2\end{cases}$

$\Rightarrow DB^2+BM^2=MD^2$

$\Rightarrow \widehat{DBM}=90^o$

$\widehat{BMC}=180^o-(\widehat{AMC}+\widehat{AMB})=180^o-(\widehat{ADB}+\widehat{AMB})=\widehat{MAD}+ \widehat{MBD}=90^o+60^o=150^o$

Anh mod sửa cả tiêu đề bài đi
 
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