toán

C

congchuaanhsang

Kẻ đường cao AH

$\Delta$AMC cân ở A\RightarrowMH=CH=$\dfrac{1}{2}CM$=$\dfrac{1}{4}BC$

\RightarrowBH=$\dfrac{3}{4}BC$\Rightarrow$BH=3CH$

\Leftrightarrow$\dfrac{BH}{AH}=3\dfrac{CH}{AH}$ \Leftrightarrow $\dfrac{AH}{BH}$=$\dfrac{1}{3}.\dfrac{AH}{CH}$

\Leftrightarrow$tanB=\dfrac{1}{3}tanC$
 
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