Finding the values of $Q$ from $180^{\circ}$ to $360^{\circ}$

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?

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Yes. But grownups use radians. – Gerry Myerson Jul 22 '12 at 10:34
Correct and well deduced. – DonAntonio Jul 22 '12 at 10:34
I encourage you to answer your own question, and accept the answer. Some here like to keep the number of listed 'unanswered questions' low. – mixedmath Jul 22 '12 at 13:40

You are correct. There are two values within a circle that have a given cosine (except -1,0,1). For $\cos Q=0.5$, these are $150^\circ$ and $240^\circ$ and you have found the right one.