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anonymous
 one year ago
Given the following triangle, solve for x.
anonymous
 one year ago
Given the following triangle, solve for x.

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jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}\)dw:1438558588866:dw which of the SOH CAH TOA identities, include only the angle the adjacent side the hypotenuse ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so confused, i'm sorry

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well... have you covered \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}\) yet?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well.. notice, you ARE GIVEN the hypotenuse, the adjacent side, and the angle that is just the variable "x" so, you'd want to use the identity that contains the hypotenuse, the adjacent side, and the angle only that way you solve for the angle, and use the GIVEN ones

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad thus\qquad cos(x)=\cfrac{3.5}{4.6} \\ \quad \\ cos^{1}[cos(x)]=cos^{1}\left( \cfrac{3.5}{4.6}\right)\impliedby taking\quad cos^{1}\textit{ on both sides} \\ \quad \\ \measuredangle x=cos^{1}\left( \cfrac{3.5}{4.6}\right)\)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1keep in mind that \(\bf cos^{1} [cos(whatever)]=whatever \\ \quad \\ sin^{1} [sin(whatever)]=whatever \\ \quad \\ tan^{1} [tan(whatever)]=whatever\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 0.760 but that isn't any of the answer choices...

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1the choices show angles in degrees, as opposed to "radians" so, when taking the inverse cosine, make sure your calculator is in "degree" mode, not "radian" mode
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