Let ABC be an acute-angled triangle and let D | vedclass

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(D) Given,in $	riangle ABC$,$D$ is the mid-point of $BC$ and $AB = AD$.Since $AB = AD$,we have $\angle B = \angle ADB$.Let $\angle ADB = B$. Then $\a

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$3$

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(D) Given,in $	riangle ABC$,$D$ is the mid-point of $BC$ and $AB = AD$.Since $AB = AD$,we have $\angle B = \angle ADB$.Let $\angle ADB = B$. Then $\angle ADC = \pi - B$. Let $\angle ADC = 	heta$,so $	heta = \pi - B$.Since $D$ is the mid-point of $BC$,$BD = DC = 1$ (taking ratio $m:n = 1:1$).Applying the cotangent theorem in $	riangle ABC$ with cevian $AD$:$(m+n) \cot 	heta = n \cot B - m \cot C$Substituting $m=1, n=1$ and $	heta = \pi - B$:$(1+1) \cot(\pi - B) = 1 \cdot \cot B - 1 \cdot \cot C$$2(-\cot B) = \cot B - \cot C$$-2 \cot B = \cot B - \cot C$$\cot C = 3 \cot B$$\frac{1}{	an C} = \frac{3}{	an B}$$\frac{	an B}{	an C} = 3$.

Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB = AD$,then $\tan(B) / \tan(C)$ equals

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