If the perpendicular AD divides the base of t | vedclass

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(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$. Since $AD \perp BC$,we have $	an \alpha = \frac{BD}{AD} = \frac{2}{6} = \frac{1}{3}$ and $	a

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$\frac{\pi}{4}$

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(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$. Since $AD \perp BC$,we have $	an \alpha = \frac{BD}{AD} = \frac{2}{6} = \frac{1}{3}$ and $	an \beta = \frac{CD}{AD} = \frac{3}{6} = \frac{1}{2}$.We need to find $\angle A = \alpha + \beta$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we get:$	an A = \frac{\frac{1}{3} + \frac{1}{2}}{1 - (\frac{1}{3} 	imes \frac{1}{2})} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$.Since $	an A = 1$,we have $A = \frac{\pi}{4}$.

If the perpendicular $AD$ divides the base of the triangle $ABC$ such that $BD, CD$ and $AD$ are in the ratio $2:3:6$,then angle $A$ is equal to

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