An electric pole PN is such that PN=12cm where N is the top of the pole and P the base .At a given moment of the day the shadow of the pole PN′ = PN. find a) the length NN′ b) the bearing of P from N.


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Question Number 41290 by Rio Michael last updated on 04/Aug/18

$A n e l e c t r i c p o l e P N i s s u c h t h a t P N = 12 c m w h e r e N i s t h e t o p o f t h e p o l e a n d P t h e b a s e$ $. A t a g i v e n m o m e n t o f t h e d a y t h e s h a d o w o f t h e p o l e {P N}^{'} = P N . f i n d$ $a) t h e l e n g t h {N N}^{'}$ $b) t h e b e a r i n g o f P f r o m N .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

	$p o l e = P N$ $s h a d o w = {P N}^{'}$ $g i v e n P N = {P N}^{'} = 12$ $s o {N N}^{'} = \sqrt{12^{2} + 12^{2}} = 12 \sqrt{2}$ $w h a t i s m e a n i n g o f b e a r i n g o f P f r o m N$
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