obtuse triangle help

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volleyball_chick

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Ok, I'm trying to figure out how to find the angles of an obtuse triangle when I have all three sides. The sides are 4, 10, and 12. I don't know what to try... my first thought was the trig ratios, but in my notes I have that they're only for right triangles. If anyone can help, that'd be great!!!
 
You could start by using the law of cosines and solving for an unknown angle.

\(\displaystyle \L\\A=cos^{-1}\left(\frac{b^{2}+c^{2}-a^{2}}{2bc}\right)\)

Where a,b,c, are your side lengths and A is angle A.

Once you have an angle you can find the others by law of sines, etc.

triln0.gif
 
Thanks! And would the law of sines be the easiest way to figure it out? I don't know why, but this is really difficult for me to figure out.
 
It will be better if you use law of cosines (formula given by galactus) for all the angles.

This will avoid any ambiguity in case of obtuse angle (since cosine changes sign - from acute to obtuse).
 
Subhotosh Khan said:
It will be better if you use law of cosines (formula given by galactus) for all the angles.
Click to expand...
GOTCHA Subby: for any 2 angles; 3rd angle = 180 - (1st + 2nd) :wink:
 
What I meant was:

After calculating A with cosine function (assuming it is one of the acute angles), if C is calculated from:

sin C = c/a * sin A

we need to consider two values of C ( angle and the supplementary angle).

Then apply the summation of angles law to find the third angle and throw away one of the Cs as impossible case. Just a little more indirect.

I probably should have said "...the next angle should be calculated through law of cosines also..."
 
YES YES...I was joking: ONLY 2 angles need to be calculated.
 
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