anonymous
  • anonymous
Solve for x. Round your answer to 2 decimal places. (picture below) ** medal & fan
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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Jamierox4ev3r
  • Jamierox4ev3r
do you know what the law of sines is?
anonymous
  • anonymous
um, i dont think i have covered that yet.

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Jamierox4ev3r
  • Jamierox4ev3r
Does this look familiar to you at all? \(\large\frac{a}{Sin A}=\large\frac{b}{Sin B}=\large\frac{c}{Sin C}\)
anonymous
  • anonymous
no, i definitely have not covered that yet :-(
Jamierox4ev3r
  • Jamierox4ev3r
fair enough. I'm not sure how I would solve a problem like this without that, sorry
jdoe0001
  • jdoe0001
|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?
anonymous
  • anonymous
cosine
jdoe0001
  • jdoe0001
yeap thus... one sec
anonymous
  • anonymous
i will wait for you :-)
jdoe0001
  • jdoe0001
\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(48^o)} \)
jdoe0001
  • jdoe0001
hmmm my 58 turned into a 48 for whatever reason =) \(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(58^o)}\)
jdoe0001
  • jdoe0001
notice, you're using degrees, thus, make sure your calculator is in Degree mode when getting the cosine
anonymous
  • anonymous
x = 17 / 0.53 ?
jdoe0001
  • jdoe0001
yeap
anonymous
  • anonymous
i got it!! thank you so much!! that was really simple!!

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