# How would one calculate the cosine of an obtuse angle?

How would you calculate the cosine of an obtuse triangle's largest angle?

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"How would you calculate" is not clear enough. You did post a diagram, which is helpful. But you did not indicate what is known about the triangle. Is it the three sides? If that is the case, say so, and give the measurements. Is it one angle and two sides? The more specific you are, the more quick and relevant will be the response. – André Nicolas Jun 15 '11 at 18:42

You would use the Law of cosines

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I just noticed that you already know the law of cosines (or should know it, according to your other question)! cos = adj/hyp is the rule for right triangles, as Ross has mentioned. – The Chaz 2.0 Jun 15 '11 at 18:20
yes yes yes. but which of the two sides is the adj one? – Daniel Pendergast Jun 15 '11 at 18:31
Neither! You will use the law of sines to find the third side, if necessary, and then the law of cosines to find the cosine.... – The Chaz 2.0 Jun 15 '11 at 18:33
oh! really?? kool. thanks! – Daniel Pendergast Jun 15 '11 at 18:35
@Dan: My pleasure. Welcome to the site! – The Chaz 2.0 Jun 15 '11 at 18:38

Cosine=adjacent/hypotenuse is only true in right triangles, as that is the only time there is a hypotenuse. $\cos \theta$ is well defined for $\theta \gt 90^{\circ}$ and we have $\cos \theta=-\cos (180^{\circ}-\theta)$.

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$\cos A= -\cos(180-A)$ e.g $\cos(120)= -cos(180-120)= -\cos60$ i.e $-1/2$ or $-0.5$

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This could work if you knew the measure of angle $A$, but presumably that was not the case the OP had in mind when presenting the illustration of a triangle. See the Accepted Answer for a more likely interpretation. – hardmath Sep 11 '15 at 14:57