.
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!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. I checked your calculations for cosines and confirm that they are correct.
2. Now calculate . (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
= . (See the lesson Trigonometric functions of half argument in this site).
You will get = = = = = = .
3. Hence, the angle is congruent to the angle C. (They both are acute.)
4. Proved.
I didn't know this fact before. It is new for me. Thanks !