Question

In triangle ABC, AB = 6cm, BC = 13cm and CA = 9cm, how do you find the value of cos BAC as a fraction in its lowest terms?

Answer 1

`cos A = {b^2 + c^2 - a^2}/{2 b c} = {9^2 + 6^2 - 13^2}/{2(9)(6)} = -13/27`

Explanation:

`a^2 = b^2 + c^2 - 2 b c cos A`

`2bc cos A = b^2 + c^2 - a^2`

`cos A = {b^2 + c^2 - a^2}/{2 b c}`

`cos A = {9^2 + 6^2 - 13^2}/{2(9)(6)} = -13/27`