Triangle: Longest Side Opposite Greatest AngleDate: 10/23/1999 at 08:21:13 From: Michael Subject: Logic Dear Dr. Maths, Could you kindly help me with the following question: Prove that in any triangle, the greatest side is opposite the greatest angle. (Prove by contradiction.) Your help is appreciated. Thanking you in advance, Michael Date: 10/25/1999 at 02:18:18 From: Doctor Floor Subject: Re: Logic Hi Michael, Thanks for writing. Let a, b and c be the sides of a triangle opposite to vertices A, B, and C respectively. We will use the Law of Sines: a / sin A = b / sin B = c / sin C When angles A > B > C, then also sin A > sin B > sin C. This is easy to see when triangle ABC is acute or right (because the graph of f(x) = sin x is increasing for 0 deg <= x <= 90 deg). When triangle ABC is obtuse, then A > 90 degrees, so note that sin A = sin (180-A), and that 180-A > B as well as 180-A > C (since A+B+C = 180 degrees). So we give names to the vertices in such a way that A > B > C. So, suppose that a < b. From the Law of sines we have a = b * sin A / sin B So a < b gives sin A / sin B < 1 and sin A < sin B, and we have a contradiction. The assumption a < c leads to a contradiction in the same way. so we have proven that a is the greatest side by contradiction. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/