anonymous
  • anonymous
Given the following triangle, solve for x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jdoe0001
  • jdoe0001
well \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}\)|dw:1438558588866:dw| which of the SOH CAH TOA identities, include only the angle the adjacent side the hypotenuse ?
anonymous
  • anonymous
I'm so confused, i'm sorry
jdoe0001
  • jdoe0001
well... 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?

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anonymous
  • anonymous
yes, I have.
jdoe0001
  • jdoe0001
well.. 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
  • jdoe0001
\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad thus\qquad cos(x)=\cfrac{3.5}{4.6} \\ \quad \\ cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\impliedby taking\quad cos^{-1}\textit{ on both sides} \\ \quad \\ \measuredangle x=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\)
jdoe0001
  • jdoe0001
keep in mind that \(\bf cos^{-1} [cos(whatever)]=whatever \\ \quad \\ sin^{-1} [sin(whatever)]=whatever \\ \quad \\ tan^{-1} [tan(whatever)]=whatever\)
anonymous
  • anonymous
I got 0.760 but that isn't any of the answer choices...
jdoe0001
  • jdoe0001
the 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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