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

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

$${An}\:{electric}\:{pole}\:{PN}\:{is}\:{such}\:{that}\:{PN}=\mathrm{12}{cm}\:{where}\:{N}\:{is}\:{the}\:{top}\:{of}\:{the}\:{pole}\:{and}\:{P}\:{the}\:{base} \\ $$$$.{At}\:{a}\:{given}\:{moment}\:{of}\:{the}\:{day}\:{the}\:\boldsymbol{{shadow}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{pole}}\:{PN}'\:=\:{PN}.\:{find}\: \\ $$$$\left.{a}\right)\:{the}\:{length}\:{NN}' \\ $$$$\left.{b}\right)\:{the}\:{bearing}\:{of}\:{P}\:{from}\:{N}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

$${pole}={PN} \\ $$$${shadow}={PN}' \\ $$$${given}\:{PN}={PN}'=\mathrm{12} \\ $$$${so}\:{NN}'=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }\:=\mathrm{12}\sqrt{\mathrm{2}}\: \\ $$$${what}\:{is}\:{meaning}\:{of}\:{bearing}\:{of}\:{P}\:{from}\:{N} \\ $$

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