A vertical pole subtends an angle tan ^{-1}le | vedclass

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(C) Let the pole be $AC$ of height $h$ and $P$ be a point on the ground at a distance $x$ from the base $A$. Let $B$ be the midpoint of $AC$,so $AB =

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$\left(\frac{2}{9}, \frac{1}{4}\right)$

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(C) Let the pole be $AC$ of height $h$ and $P$ be a point on the ground at a distance $x$ from the base $A$. Let $B$ be the midpoint of $AC$,so $AB = BC = h/2$.Let $\angle APB = \beta$ and $\angle BPC = \alpha$. Then $\angle APC = \alpha + \beta = 	an^{-1}(1/2)$.In $	riangle APB$,$	an \beta = \frac{AB}{AP} = \frac{h/2}{x} = \frac{h}{2x}$.In $	riangle APC$,$	an(\alpha + \beta) = \frac{AC}{AP} = \frac{h}{x} = 2 	an \beta = 1/2$,so $	an \beta = 1/4$.Since $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = 1/2$,substitute $	an \beta = 1/4$:$\frac{	an \alpha + 1/4}{1 - 	an \alpha / 4} = 1/2$$2(	an \alpha + 1/4) = 1 - 	an \alpha / 4$$2 	an \alpha + 1/2 = 1 - 	an \alpha / 4$$2 	an \alpha + 	an \alpha / 4 = 1 - 1/2$$\frac{9}{4} 	an \alpha = 1/2$$	an \alpha = \frac{1}{2} 	imes \frac{4}{9} = 2/9$.Thus,$(	an \alpha, 	an \beta) = (2/9, 1/4)$.

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$,then $(\tan \alpha, \tan \beta)$ is equal to

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