In triangle ABC, we have AB = 5, BC = 8, and AC = 9, as shown below

Calculate cos A, cos B, cos C.

Guest Feb 7, 2022

#1**0 **

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!

BuilderBoi Feb 8, 2022

#2**+1 **

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 .....

ElectricPavlov Feb 8, 2022