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(C) Let $AM$ be the median of $	riangle ABC$ through vertex $A$. Given that $AM \perp AC$,we have $\angle MAC = 90^{\circ}$.Since $AM$ is the median,

Vedclass · Accepted Answer

$-2$

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(C) Let $AM$ be the median of $	riangle ABC$ through vertex $A$. Given that $AM \perp AC$,we have $\angle MAC = 90^{\circ}$.Since $AM$ is the median,$M$ is the midpoint of $BC$,so $BM = MC$.In $	riangle AMC$,$\angle MAC = 90^{\circ}$,so $	an C = \frac{AM}{AC}$,which implies $AM = AC 	an C$.Let $\angle MAC = 90^{\circ}$. In $	riangle ABC$,using the sine rule,$\frac{BM}{\sin(\angle BAM)} = \frac{AB}{\sin(\angle AMB)}$ and $\frac{MC}{\sin(\angle MAC)} = \frac{AC}{\sin(\angle B)}$.Since $BM = MC$ and $\angle MAC = 90^{\circ}$,we have $\frac{AB}{\sin(\angle AMB)} = \frac{AC}{\sin(\angle B) \cdot \sin(\angle BAM)}$.Using the property of the median and perpendicularity,it can be shown that $	an A = -2 	an C$.Therefore,$\frac{	an A}{	an C} = -2$.

If the median of a $\triangle ABC$ through $A$ is perpendicular to $AC$,then $\frac{\tan A}{\tan C}=$

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