<p>Alright so theres an isoceles triangle with sides a=9, b= 13, and c=17. There is one angle given which A=31.47. The question wants me to find angle C. When i use the law of sines i get C=80.435, and when i use the law of cosines i get C=99.59. Which angle is right and why?</p>
<p>ummmm isn't the definition of an isoceles triangle having two sides/angles that are the same measures?</p>
<p>owned.....</p>
<p>well then forget about the isoceles part, i just added that in there on my own stupidity, i suck at classifying triangles and crap. Please help im confused</p>
<p>Could you draw it out for us?</p>
<p>On Paint or something. I'm having a hard time placing angle A.</p>
<p>B </p>
<p>--------C----------------------A </p>
<p>if you connect the letters you will get the shape of the triangle. At first i thought to use the law of cosines, since it gives all three sides, but when i do the law of sines it condradicts the answer, maybe this is an ambigous case.</p>
<p>Sin(A)/a= Sin(C)/c plug all of your variables in to solve for C.</p>
<p>bump (10 chars)</p>
<p>islanders08 just answered the question...use The law of sines and you get C... check out this website too... <a href="http://www.analyzemath.com/Triangle/SineLaw.html%5B/url%5D">http://www.analyzemath.com/Triangle/SineLaw.html</a></p>
<p>sin(31.47)/9 = sin(C)/17
Sin(C) = (17sin(31.47))/9
C = arcsin(...) = 80.435
180 - 80.435 = 99.594</p>
<p>Cosines...</p>
<p>17^2 = 9^2 + 13^2 - 2(9)(13)cos(C)
39 = -234cos(C)
C = arccos(-39/234) = 99.594</p>
<p>I think the law of cosines is correct in this case. The law of sines doesn't take the side b into account, so from the information you use with it (side, side, angle) there could be two possible triangles. If you subtract the angle given by the law of sines from 180, you get the correct angle, 99.594. With the law of cosines, given three sides, there is only one possible triangle. I suck at explaining things like this but I hope you understand what I'm saying.</p>
<p>thanks sean, you definetly answered my question.</p>