[TOÁN 9] tslg

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thaonguyenkmhd

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Do $AC=AM \to \Delta ACM$ cân tại A nên đường cao AH đồng thời là trung tuyến $\to$ CH=HM

Ta có $BM=CM=2CH \to BM+HM=2CH+HM \to BH=3CH$

$\Delta ACH(\hat{H}=90^o): tanC=\dfrac{AH}{CH} \\ \Delta ABH(\hat{H}=90^o): tanB=\dfrac{AH}{BH}=\dfrac{AH}{3CH} \\ \to tanB=\dfrac{1}{3}tanC$
 
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