+382 Given the lengths of the sides of a triangle are in the ratio of 3:8:6, what is the degree measure of the largest angle? Please explain.
+129852 We can use the Law of Cosines to solve this [although there may be other ways, too]
We have that
8^2 = 3^2 + 6^2 - 2 (3)(6) cos A rearrange as
[8^2 - 3^2 -6^2] / [ -2 (3) (6) ] = cos A simplify
[ 19] / [ -36] = cos A
-19/36 = cos A and we can find the angle A with the cosine inverse (arccos)
arccos ( -19/36) = A ≈ 121.86°
P.S. - this angle is obtuse ( > 90°)....and a triangle may only have one angle > 90°...so...we know it's the largest
