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anonymous
 one year ago
Help with a trigonometric question, please! It's posted in the comments.
anonymous
 one year ago
Help with a trigonometric question, please! It's posted in the comments.

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jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.2\(\bf 0<\theta<90^o\textit{means the 1st quadrant, means sine and cosine are both positive} \\ \quad \\ sin(\theta)=\cfrac{21}{29}\qquad sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad thus \\ \quad \\ sin(\theta)=\cfrac{21}{29}\to \cfrac{opposite}{hypotenuse}\to \cfrac{b}{c}\to \cfrac{b=21}{c=29} \\ \quad \\ \textit{using the pythagorean theorem }\to c^2=a^2+b^2\implies \pm\sqrt{c^2b^2}=a \\ \quad \\ \pm\sqrt{29^221^2}=a\) what would that give youu for "a" then? recall "a" = "adjacent side"dw:1435102900196:dw

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.2yeap so now we know that the adjacent side is 20 notice that, the root could give a 20 or a +20, so it couild really be either because either one squared, will give 400 anyway BUT, we know the angle is in the 1st quadrant, and on the 1st quadrant, cosine is positive and also is the adjacent side, thus we'll use the +20 then thus \(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\to \cfrac{a}{c}\to \cfrac{20}{29}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is C. Thanks! It was really easy to follow you and understand what you were trying to explain.
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