Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

if you must find the Angle C based on the sides of a = 2, 3 b = 4,6 og c = 5, 9 

I have used the formula:

$$\cos (C) =\frac{a^2 + b^2-c^2}{2ab}$$

use, but I think i'm doing something wrong:

$$\cos(C) = \frac{(2,3^2) + (4,6^2) - (5,9^2)}{2 \cdot 2,3 \cdot 4,6} = -0,395085066$$

$$C = \cos^{-1} ( -0,395085066 ) = \cdots$$

share|cite|improve this question
Is it $b=4.9$ (first line) or $4.6$ (where you calculate $\cos(C)$)? – André Nicolas Feb 5 '12 at 21:23
The first line define side b = 4.9 , b is already defined. The problem is " -0.395085066 " – user1022734 Feb 5 '12 at 21:27
Nothing wrong with $-0.39508\dots$. Sure, it is negative, but obtuse angles have negative cosines. – André Nicolas Feb 5 '12 at 21:45
Ah, my bad then – user1022734 Feb 5 '12 at 21:50
up vote 4 down vote accepted

By the Cosine Law, we have $$c^2=a^2+b^2-2ab\cos(C).$$ This can be rewritten as $$\cos(C)=\frac{a^2+b^2-c^2}{2ab}.$$

Put $a=2.3$, $b=4.6$, and $c=5.9$. We get $\cos(C)=-0.395051$. So far, we agree.

Pressing the $\cos^{-1}$ button, we get $C=113.27128$ (degrees). Doing it in radian mode, we get $C=1.9769568$ (radians). But probably you want the answer in degrees.

You do not explain why you think the answer is not correct. I noticed that in your first line you say that $b=4.9$, but in your calculation you write $4.6$, and indeed you used $4.6$. If $b$ was really supposed to be $4.9$, you should redo the calculation using $b=4.9$.

One other possibility is that you do not know how to find $C$, knowing its cosine and the fact that the angle is between $0$ and $180$. The notation is a little confusing. Note that $\cos^{-1}x$ does not mean $\frac{1}{\cos x}$.

In a comment, you mention that the issue is with "$-0.395051$", but do not explain why you see it an issue. I will guess that it is the fact that the number is negative. No problem! The cosine of any obtuse angle is negative. The reason we got a negative is that $c^2>a^2+b^2$. In that case, the angle $C$ will always be obtuse.

share|cite|improve this answer
Thanks, just wanted some confirm! – user1022734 Feb 5 '12 at 21:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.