Question 600414
Well assuming you have a right triangle ABC where B is the right angle, you can have two possible triangles: 


AB = 3, BC = 4, AC = 5


or


AB = 4, BC = 3, AC = 5



The reason why AC is stuck at 5 is because the hypotenuse is always the longest segment of any right triangle. 




So 


cos(A) = adjacent/hypotenuse = AB/AC = 3/5


which means cos(A) = 3/5


or


cos(A) = adjacent/hypotenuse = AB/AC = 4/5


which means cos(A) = 4/5


and that's about it. 



So either there's a typo and you only need to find two possible values of cos(A) or this triangle isn't a right triangle.



Let's assume that there isn't a typo. This would mean that this isn't a right triangle. But this has to be a right triangle because these three values are a Pythagorean triple (also, {{{3^2+4^2=5^2}}}). So that's why I'm convinced that there's a typo.