AB is a vertical pole resting at the end A on | vedclass

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(B) Let $AC = x = CB$. Then $AB = AC + CB = 2x$.Given $AP = 3 \, AB = 3(2x) = 6x$.Let $\angle APC = \alpha$. In $\Delta ACP$,$	an \alpha = \frac{AC}{

Vedclass · Accepted Answer

$\frac{3}{19}$

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(B) Let $AC = x = CB$. Then $AB = AC + CB = 2x$.Given $AP = 3 \, AB = 3(2x) = 6x$.Let $\angle APC = \alpha$. In $\Delta ACP$,$	an \alpha = \frac{AC}{AP} = \frac{x}{6x} = \frac{1}{6}$.In $\Delta ABP$,$\angle APB = \alpha + \beta$. Thus,$	an(\alpha + \beta) = \frac{AB}{AP} = \frac{2x}{6x} = \frac{1}{3}$.Using the formula $	an \beta = 	an((\alpha + \beta) - \alpha) = \frac{	an(\alpha + \beta) - 	an \alpha}{1 + 	an(\alpha + \beta)	an \alpha}$.Substituting the values: $	an \beta = \frac{\frac{1}{3} - \frac{1}{6}}{1 + (\frac{1}{3})(\frac{1}{6})} = \frac{\frac{2-1}{6}}{1 + \frac{1}{18}} = \frac{\frac{1}{6}}{\frac{19}{18}} = \frac{1}{6} 	imes \frac{18}{19} = \frac{3}{19}$.

$AB$ is a vertical pole resting at the end $A$ on the level ground. $P$ is a point on the level ground such that $AP = 3 \, AB$. If $C$ is the mid-point of $AB$ and $CB$ subtends an angle $\beta$ at $P$,the value of $\tan \beta$ is

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