anonymous
  • anonymous
PLEAAASSEEE help Pre Calc: thank you so much picture posted below xoxo vvv
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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jim_thompson5910
  • jim_thompson5910
you first have to use the pythagorean theorem to find the length of AB
jim_thompson5910
  • jim_thompson5910
AC^2 + BC^2 = AB^2 36^2 + 24^2 = x^2 1296 + 576 = x^2 keep going to solve for x

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anonymous
  • anonymous
x=+/- 12 to the power of 13
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
you mean \[\Large x = \pm 12\sqrt{13}\] right?
anonymous
  • anonymous
yes :) @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
since AB is the length and x is the length, we only consider positive lengths so \[\Large x = 12\sqrt{13}\] which means \[\Large AB = 12\sqrt{13}\]
jim_thompson5910
  • jim_thompson5910
now use the definition of sine, plug in the lengths, and simplify sin(angle) = opposite/hypotenuse sin(A) = 24/(12*sqrt(13)) sin(A) = 2/(sqrt(13)) sin(A) = (2*sqrt(13))/(sqrt(13)*sqrt(13)) sin(A) = (2*sqrt(13))/(13)
anonymous
  • anonymous
ughhh thank you so much! FINALLY I understand this nonsense.
jim_thompson5910
  • jim_thompson5910
that's great

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