SOLUTION: given a triangle with sides 4, 5, 6. find the EXACT values of the angles to show that one angle is EXACTLY twice the other (no estimations).

Algebra ->  Trigonometry-basics -> SOLUTION: given a triangle with sides 4, 5, 6. find the EXACT values of the angles to show that one angle is EXACTLY twice the other (no estimations).      Log On


   

Question 275040: given a triangle with sides 4, 5, 6. find the EXACT values of the angles to show that one angle is EXACTLY twice the other (no estimations).
Answer by kensson(21) About Me  (Show Source):
You can put this solution on YOUR website!
Interesting!
The cosine rule is: +a%5E2+=+b%5E2+%2B+c%5E2+-+2bc+cos+A+
Or, rearranging, cos+A+=+%28b%5E2+%2B+c%5E2+-+a%5E2%29%2F%282bc%29
So the cosine of the angle opposite the 4 side is %285%5E2+%2B+6%5E2+-+4%5E2%29%2F%282%2A5%2A6%29 = 45/60 = 3/4.
Opposite the 5: %284%5E2+%2B+6%5E2+-+5%5E2%29%2F%282%2A4%2A6%29 = 27/48 = 9/16.
Opposite the 6: %284%5E2+%2B+5%5E2+-+6%5E2%29%2F%282%2A4%2A5%29 = 5/40 = 1/8.
So the smallest angle is arccos(3/4), the middle angle is arccos(9/16) and the biggest angle arccos(1/8).
Now, cos+2x+=+2+cos%5E2%28x%29+-+1. For the smallest angle:
cos+2x+=+2+%283%2F4%29%5E2+-1+=+2+%2A+9%2F16+-+1+=+1%2F8
So the largest angle is twice the size of the smallest one.