You are watching: Find the height of the flagpole

I"m confused about the given problem over because ns can"t recognize yet the length of the shadow, and also I"m stuck at the edge of depression, in i m sorry I an alleged that it needs to be the angle of elevation instead.

My solution:

$\sin\left(58^\circ \right)=\fracxx-3$

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trigonometry

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edited jan 22 "18 in ~ 22:08

Xtravagant

asked jan 22 "18 at 21:42

XtravagantXtravagant

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First, we draw a flagpole, $x$m long. Climate we draw a shadow that is $x-3$m long. Then the sunlight ray i beg your pardon casts the zero connects the optimal of the pole and TOP that the shadow. This method the angle developed by the hypotenuse (the sunlight ray) and the zero is $58^o$. Now we have $\tan 58^o=\fracxx-3$, offering us $x\tan 58^o-3\tan 58^o=x$. We lug $x$ to one side, having $x\left(\tan 58^o-1\right)=3\tan 58^o$. $x=7.997...=8.00\left(3sf\right)$.

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edited Feb 19 "20 in ~ 8:48

wuyudi

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answered jan 22 "18 at 21:52

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