Triangle: Longest Side Opposite Greatest Angle

Date: 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/