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(B) Let $AM$ be the median through $A$ to the side $BC$. Given $AM \perp AB$,so $\angle BAM = 90^\circ$.In $	riangle ABM$,$\angle B + \angle BAM + \a

Vedclass · Accepted Answer

$2$

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(B) Let $AM$ be the median through $A$ to the side $BC$. Given $AM \perp AB$,so $\angle BAM = 90^\circ$.In $	riangle ABM$,$\angle B + \angle BAM + \angle AMB = 180^\circ \implies \angle AMB = 180^\circ - 90^\circ - B = 90^\circ - B$.Since $AM$ is a median,$BM = MC = \frac{a}{2}$.In $	riangle ABM$,$	an B = \frac{AM}{BM} = \frac{AM}{a/2} \implies AM = \frac{a}{2} 	an B$.In $	riangle AMC$,$\angle AMC = 180^\circ - (90^\circ - B) = 90^\circ + B$.Using the sine rule in $	riangle AMC$: $\frac{AM}{\sin C} = \frac{MC}{\sin \angle MAC}$.Also,$\angle MAC = A - 90^\circ$.So,$\frac{AM}{\sin C} = \frac{a/2}{\sin(A - 90^\circ)} = \frac{a/2}{-\cos A}$.Substituting $AM = \frac{a}{2} 	an B$: $\frac{\frac{a}{2} 	an B}{\sin C} = \frac{a/2}{-\cos A} \implies \frac{	an B}{\sin C} = \frac{-1}{\cos A}$.Using $\sin C = \sin(180^\circ - (A+B)) = \sin(A+B) = \sin A \cos B + \cos A \sin B$.$	an B (-\cos A) = \sin A \cos B + \cos A \sin B$.$-\frac{\sin B}{\cos B} \cos A = \sin A \cos B + \cos A \sin B$.$-\sin B \cos A = \sin A \cos^2 B + \cos A \sin B \cos B$.Dividing by $\cos A \cos B$: $-	an B = 	an A \cos B + 	an B \cos B$ (This approach is complex).Alternatively,using the cotangent theorem on $	riangle ABC$ with cevian $AM$ dividing $BC$ in ratio $1:1$:$(m+n) \cot 	heta = m \cot \alpha - n \cot \beta \implies (1+1) \cot(90^\circ - B) = 1 \cot(90^\circ) - 1 \cot(A-90^\circ)$.$2 	an B = 0 - (-	an A) = 	an A$.Therefore,$\frac{	an A}{	an B} = 2$.

If the median of a triangle $ABC$ through $A$ is perpendicular to $AB$,then the value of $\frac{\tan A}{\tan B}$ is equal to

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