+25 In a triangle ABC where AB = 6.71cm BC = 5.83cm and AC = 7.81cm . What is angle CAB?
Use the law of cosines:
c^2 = a^2 + b^2 - 2ab cos C where a, b, c are your sides, and C is your angle.
Your problem should then look like:
5.83^2 = 6.71^2 + 7.81^2 - 2(6.71)(7.81)cos C
(Use 5.83 on the left because that side is across from your angle.)
Now solve for cos C:
cos C = (5.83^2 - 6,71^2 - 7.81^2) / (-2 * 6.71 * 7.81)
cos C = 0.687255
Using arccos you then get approximately 46.6 degrees.
