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

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Question

(C) Let the pole be $AC$ of height $2h$,where $B$ is the midpoint such that $AB = BC = h$. Let $PA = x$. The angle subtended by the whole pole is $	h

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the pole be $AC$ of height $2h$,where $B$ is the midpoint such that $AB = BC = h$. Let $PA = x$. The angle subtended by the whole pole is $	heta = 	an^{-1}\left(\frac{1}{2}ight)$,so $	an 	heta = \frac{2h}{x} = \frac{1}{2}$,which implies $x = 4h$.Now,$	an(	heta + \beta) = \frac{2h}{x} = \frac{1}{2}$ and $	an \beta = \frac{h}{x} = \frac{h}{4h} = \frac{1}{4}$.Since $	heta = \alpha + \beta$,we have $\alpha = 	heta - \beta$.Using the formula $	an \alpha = 	an(	heta - \beta) = \frac{	an 	heta - 	an \beta}{1 + 	an 	heta 	an \beta} = \frac{\frac{1}{2} - \frac{1}{4}}{1 + (\frac{1}{2})(\frac{1}{4})} = \frac{\frac{1}{4}}{1 + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{9}{8}} = \frac{1}{4} 	imes \frac{8}{9} = \frac{2}{9}$.Thus,$(	an \alpha, 	an \beta) = \left(\frac{2}{9}, \frac{1}{4}ight)$.

$A$ 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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