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anonymous

  • one year ago

Solve for x. Round your answer to 2 decimal places. (picture below) ** medal & fan

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  1. anonymous
    • one year ago
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  2. Jamierox4ev3r
    • one year ago
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    do you know what the law of sines is?

  3. anonymous
    • one year ago
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    um, i dont think i have covered that yet.

  4. Jamierox4ev3r
    • one year ago
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    Does this look familiar to you at all? \(\large\frac{a}{Sin A}=\large\frac{b}{Sin B}=\large\frac{c}{Sin C}\)

  5. anonymous
    • one year ago
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    no, i definitely have not covered that yet :-(

  6. Jamierox4ev3r
    • one year ago
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    fair enough. I'm not sure how I would solve a problem like this without that, sorry

  7. jdoe0001
    • one year ago
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    |dw:1440460690118:dw| recall you SOH CAH TOA \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}\) which identity uses the angle adjacent side, and hypotenuse only?

  8. anonymous
    • one year ago
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    cosine

  9. jdoe0001
    • one year ago
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    yeap thus... one sec

  10. anonymous
    • one year ago
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    i will wait for you :-)

  11. jdoe0001
    • one year ago
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    \(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(48^o)} \)

  12. jdoe0001
    • one year ago
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    hmmm my 58 turned into a 48 for whatever reason =) \(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(58^o)}\)

  13. jdoe0001
    • one year ago
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    notice, you're using degrees, thus, make sure your calculator is in Degree mode when getting the cosine

  14. anonymous
    • one year ago
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    x = 17 / 0.53 ?

  15. jdoe0001
    • one year ago
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    yeap

  16. anonymous
    • one year ago
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    i got it!! thank you so much!! that was really simple!!

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