Difference between arccos and Cos^-1

I'm trying to use inverse cosine, in the cosine rule to find an angle of a triange when you know 3 sides. I know this formula and have it written down. However I've left my calculator and I'm having to use Google:

http://www.google.co.uk/webhp?hl=en#sclient=psy&hl=en&site=webhp&source=hp&q=arccos((5.9%5E2+%2B+4.1%5E2+-+3.2%5E2)+%2F+(2++5.9++4.1))++in+degrees&aq=f&aqi=&aql=&oq=&gs_rfai=&pbx=1&qscrl=1&fp=3dd9e71350957573

but it doesn't return the correct answer which makes think theres a difference between arccos and inverse cos.

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1 Answer

Wolfram Alpha gives about 31.21 degrees. My calculator gives 31.20576 When I first pasted it into Google, it didn't bring up the calculator, then when I put it into Alpha it lost the s off arccos. Also between the numbers in the denominator I get spaces instead of asterisks, so maybe it is not multiplying.

Some people distinguish between the inverse cosine function (which has the range restricted so there is only one value) and arccos not having the range restricted so it is multivalued, but that doesn't seem to be your problem.

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i thought I had this nailed but does it matter which value goes into a b or c (obviously each letter has to be the same throughout)? the formula I have is: – Jonathan. Nov 1 '10 at 21:53
cos^-1((b^2 + c^2 - a^2)/(2bc)) = angle – Jonathan. Nov 1 '10 at 21:53
Your formula is correct. When you choose a, b, and c that designates which angle you are solving for. In your previous comment you are solving for the angle across from side a. I think your only problem was getting the formula into the calculator properly, particularly the multiplies in the denominator. – Ross Millikan Nov 1 '10 at 22:02
Thanks Ross, but I realized that 'a' had to be the side opposite the angle I'm trying to find, which you wouldn't have known I got wrong as I didn't give the triangle in question. – Jonathan. Nov 1 '10 at 22:02