In triangle ABC, we have AB = 5, BC = 8, and AC = 9, as shown below
Calculate cos A, cos B, cos C.
+2668 cos = \(\text{adjacent}\over\text{hypotenuse}\)
The hypotenuse is always the longest side, and adjacent is the one side that is next to the relative position of the vertice that is being "cos'ed". For example: \(\color{brown}\boxed{{\cos A} = {5\over 9}}\). You can solve the rest from here. Good Luck!
+37146 BuliderBoi....
That definition only works in RIGHT TRIANGLES
to solve this problem you will need to use the Law of Cosines
c^2 = a^2 + b^2 - 2ab cos (angle c)
For cos A
8^2 = 9^2 + 5^2 - 2(9)(5) cos A then solve for cos A The others are similar .....
