Question 1026992
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Prove that in any triangle that has the sides proportionate with 4, 5 and 6, the least angle measure is half of the biggest angle measure.
With the Cosinus Theorem, I found that(AB=4k , AC=5k, BC=6k)
cos A = 2/16
cos B = 9/16
cos C = 12/16
Help me, I dont know what to do further!
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1.  I checked your calculations for cosines and confirm that they are correct.

2. Now calculate {{{cos(A/2)}}}.   (Notice that in your notations A is the largest angle, since it is opposite to the longest side BC.)

   Use the formula for the cosines of the half argument of Trigonometry   

   {{{cos(A/2)}}} = {{{sqrt((1+cos(A))/2)}}}.         (See the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A> in this site).

   You will get  {{{cos(A/2)}}} = {{{sqrt((1 + (2/16))/2)}}} = {{{sqrt(18/(16*2))}}} = {{{sqrt(9/16)}}} = {{{3/4}}} = {{{12/16}}} = {{{cos(C)}}}.

3.  Hence, the angle  {{{A/2}}}  is congruent to the angle  C.   (They both are acute.)

4.  Proved.
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I didn't know this fact before. It is new for me.  Thanks !