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anonymous

  • one year ago

Given the following triangle, solve for x.

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  1. jdoe0001
    • one year ago
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    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 ?

  2. anonymous
    • one year ago
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    I'm so confused, i'm sorry

  3. jdoe0001
    • one year ago
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    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?

  4. anonymous
    • one year ago
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    yes, I have.

  5. jdoe0001
    • one year ago
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    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

  6. jdoe0001
    • one year ago
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    \(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad thus\qquad cos(x)=\cfrac{3.5}{4.6} \\ \quad \\ cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\impliedby taking\quad cos^{-1}\textit{ on both sides} \\ \quad \\ \measuredangle x=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\)

  7. jdoe0001
    • one year ago
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    keep in mind that \(\bf cos^{-1} [cos(whatever)]=whatever \\ \quad \\ sin^{-1} [sin(whatever)]=whatever \\ \quad \\ tan^{-1} [tan(whatever)]=whatever\)

  8. anonymous
    • one year ago
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    I got 0.760 but that isn't any of the answer choices...

  9. jdoe0001
    • one year ago
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    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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