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I"m confused about the given problem above because I can"t determine yet the length of the shadow, and also I"m stuck at the angle of depression, in which I supposed that it has to be the angle of elevation instead.

My solution:

$sinleft(58^circ ight)=fracxx-3$

enter image description here

trigonometry

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Folshort

edited Jan 22 "18 at 22:08

Xtravagant

asked Jan 22 "18 at 21:42

XtravagantXtravagant

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## 1 Answer 1

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First, we draw a flagpole, $x$m long. Then we draw a shadow that is $x-3$m long. Then the sun ray which casts the shadow connects the TOP of the pole and TOP of the shadow. This means the angle formed by the hypotenuse (the sun ray) and the shadow is $58^o$. Now we have $ an 58^o=fracxx-3$, giving us $x an 58^o-3 an 58^o=x$. We bring $x$ to one side, having $xleft( an 58^o-1 ight)=3 an 58^o$. $x=7.997...=8.00left(3sf ight)$.

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Folshort

edited Feb 19 "20 at 8:48

wuyudi

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answered Jan 22 "18 at 21:52

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