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I have $\cos{Q}=-0.5$, required to find $180^{\circ}\leq Q\leq 360^{\circ}$.

What I tried: The acute angle is $60^{\circ}$, then, since cosine is negative on third quadrant, I suppose the value is $180^{\circ}+60^{\circ}=240^{\circ}$. Am I right?

  • 2
    Yes. But grownups use radians.2012-07-22
  • 0
    Correct and well deduced.2012-07-22
  • 0
    I encourage you to answer your own question, and accept the answer. Some here like to keep the number of listed 'unanswered questions' low.2012-07-22

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