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(D) Let $\angle APC = \alpha$. In $	riangle APC$,$	an \alpha = \frac{AC}{AP}$.Since $C$ is the mid-point of $AB$,$AC = \frac{1}{2} AB$. Given $AP =

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$\frac{2}{9}$

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(D) Let $\angle APC = \alpha$. In $	riangle APC$,$	an \alpha = \frac{AC}{AP}$.Since $C$ is the mid-point of $AB$,$AC = \frac{1}{2} AB$. Given $AP = 2AB$,we have $	an \alpha = \frac{\frac{1}{2} AB}{2 AB} = \frac{1}{4}$.Now,consider $	riangle ABP$. $\angle BAP = 90^{\circ}$. $\angle BPC = \beta$ and $\angle APC = \alpha$,so $\angle BAP = \alpha + \beta$.$	an(\alpha + \beta) = \frac{AB}{AP} = \frac{AB}{2 AB} = \frac{1}{2}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we have:$\frac{1}{2} = \frac{\frac{1}{4} + 	an \beta}{1 - \frac{1}{4} 	an \beta}$.Multiplying both sides by $4(1 - \frac{1}{4} 	an \beta)$:$2(1 - \frac{1}{4} 	an \beta) = 1 + 4 	an \beta$$2 - \frac{1}{2} 	an \beta = 1 + 4 	an \beta$$1 = 4.5 	an \beta$$1 = \frac{9}{2} 	an \beta$$	an \beta = \frac{2}{9}$.

Let a vertical tower $AB$ have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that $AP = 2AB$. If $\angle BPC = \beta$,then $\tan \beta$ is equal to:

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