anonymous
  • anonymous
Draw the angle given in degrees on the unit circle where 0 radians corresponds to the positive portion of the horizontal axis: -16pi/7
Trigonometry
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chestercat
  • chestercat
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anonymous
  • anonymous
anyone ?
anonymous
  • anonymous
plz help me
anonymous
  • anonymous
plz help me

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jdoe0001
  • jdoe0001
\(\bf \cfrac{16}{7}\implies \cfrac{7+7+2}{7}\implies \cfrac{\cancel{7}}{\cancel{7}}+\cfrac{\cancel{7}}{\cancel{7}}+\cfrac{2}{7}\implies 1+1+\cfrac{2}{7} \\ \quad \\ 2+\cfrac{2}{7} \\ \quad \\ thus\qquad -\cfrac{16}{7}\implies -2\pi -\cfrac{2\pi }{7}\) notice, a full CLOCKWISE cycle is \(-2\pi\) so just draw the remaining after the cycle and then get the CLOCKWISE angle
jdoe0001
  • jdoe0001
shoot... actually that's backwards
jdoe0001
  • jdoe0001
notice, a full CLOCKWISE cycle is −2π so just draw the remaining after the cycle and then get the COUNTER-CLOCKWISE angle
jdoe0001
  • jdoe0001
anyhow \(\bf -\cfrac{16\pi }{7}\implies -2\pi -\cfrac{2\pi }{7}\)

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