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(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.Given $AP = n \cdot AB = nh$.In $	riangle PAC$,$	an(\a

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$n = (2n^2 + 1)	an \alpha$

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(D) Let $AB = h$. Since $C$ is the midpoint of $AB$,$AC = \frac{h}{2}$ and $CB = \frac{h}{2}$.Given $AP = n \cdot AB = nh$.In $	riangle PAC$,$	an(\angle APC) = \frac{AC}{AP} = \frac{h/2}{nh} = \frac{1}{2n}$.In $	riangle PAB$,$	an(\angle APB) = \frac{AB}{AP} = \frac{h}{nh} = \frac{1}{n}$.We know $\alpha = \angle APB - \angle APC$.Therefore,$	an \alpha = 	an(\angle APB - \angle APC) = \frac{	an(\angle APB) - 	an(\angle APC)}{1 + 	an(\angle APB) \cdot 	an(\angle APC)}$.Substituting the values: $	an \alpha = \frac{\frac{1}{n} - \frac{1}{2n}}{1 + (\frac{1}{n})(\frac{1}{2n})} = \frac{\frac{1}{2n}}{1 + \frac{1}{2n^2}} = \frac{\frac{1}{2n}}{\frac{2n^2 + 1}{2n^2}} = \frac{1}{2n} \cdot \frac{2n^2}{2n^2 + 1} = \frac{n}{2n^2 + 1}$.Thus,$n = (2n^2 + 1)	an \alpha$.

$AB$ is a vertical tower. The point $A$ is on the ground and $C$ is the middle point of $AB$. The part $CB$ subtends an angle $\alpha$ at a point $P$ on the ground. If $AP = n \cdot AB$,then the correct relation is:

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