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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:

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

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

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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

1 Answer 1

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.

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Thanks, just wanted some confirm! –  user1022734 Feb 5 '12 at 21:50

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