Law of Cosines in Informal Trigonometric ProofDate: 12/22/2004 at 22:01:20 From: andy Subject: determining if triangle ABC is either isosceles or right I'm in trig in high school and came across this problem involving cosine: If cosA/cosB = b/a, then show that triangle ABC is either isosceles or right. I drew a right triangle with the right angle at C. Then I labeled sides a,b,c opposite the repective angle letters; thus, side c is the hypotenuse. Now, using this right triangle, I come up with: cosA = b/c cosB = a/c Since they say cosA/cosB, I substituted cosA and cosB with b/c and a/c to get (b/c)/(a/c) which simplifies to b/a. It seems like that proves that cosA/cosB = b/a using a right triangle. However, I feel like I haven't answered the question completely and I'm not sure what to do with the isosceles triangle. Can you tell me what I'm missing? Date: 12/23/2004 at 04:05:59 From: Doctor Floor Subject: Re: determining if triangle ABC is either isosceles or right Hi Andy, Thanks for your question. We should use the Law of Cosines: a^2 = b^2 + c^2 - 2bc cosA or, after some rewriting. cosA = (-a^2 + b^2 + c^2)/(2bc) Similarly cosB = (a^2 - b^2 + c^2)/(2ac) This gives cosA a (-a^2 + b^2 + c^2) ---- = -------------------- cosB b (a^2 - b^2 + c^2) Now cosA/cosB = b/a is equivalent to b a (-a^2 + b^2 + c^2) - = -------------------- a b (a^2 - b^2 + c^2) and cross-multiplying we get a^2(-a^2 + b^2 + c^2) = b^2(a^2 - b^2 + c^2). Now, can you finish by further solving this equation? If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 12/23/2004 at 16:12:32 From: andy Subject: determining if triangle ABC is either isosceles or right Dear Dr. Floor, Thank you for the help so far. I've followed through every step on my own piece of paper and it seems like if I prove that a = b, then this equation works for an isosceles triangle, but I'm not really sure. OK, here's my work from the point you left off which was a^2(-a^2 + b^2 + c^2) = b^2(a^2 - b^2 + c^2) -a^4 + a^2b^2 + a^2c^2 = a^2b^2 - b^4 + c^2b^2 (distributed) -a^4 + a^2c^2 = -b^4 + c^2b^2 (cancel a^2b^2 on both sides) b^4 - c^2b^2 + a^2c^2 - a^4 = 0 (bring all to one side).. However, first I'm not sure exactly what I'm trying to actually solve, and second, how, when I have solved it, I will prove that this triangle is isosceles or right. I hope you can help me a little bit more. Thanks a lot! Date: 12/27/2005 at 03:08:02 From: Doctor Floor Subject: Re: determining if triangle ABC is either isosceles or right Hi Andy, Your work so far looks good. You have now found an equation that is quadratic with regard to b^2 (try replacing b^2 with x if you're not sure what I mean): b^4 - c^2b^2 + a^2c^2 - a^4 = 0 (b^2)^2 - c^2(b^2) + a^2c^2 - a^4 = 0 (x)^2 - c^2(x) + a^2c^2 - a^4 = 0 This you can solve for b^2 with the quadratic formula, where in the formula a = 1, b = -(c^2), and c = a^2c^2 - a^4. Note that those are not the same a,b,c you are working with from your triangle, those are the a,b,c that are traditionally used in the quadratic formula, defined in terms of the a,b,c from the triangle. That gives you b^2 = .... or b^2 = .... (because of the +/- in the formula) Hopefully you will find that b^2 = a^2 (isosceles triangle) or b^2 = c^2 - a^2 (right triangle). Good luck! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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