Solution
∠A is the largest; ∠C is the smallest.
Solution 1. We can use the law of cosines:
62 = 42 + 52 - 2 · 4 · 5 cos A
cos A = (62 - 42 - 52) / (-2 · 4 · 5) = 0.125
∠A = arccos (0.125) ≈ 82.82°
42 = 52 + 62 - 2 · 5 · 6 cos C
cos C = (42 - 52 - 62) / (-2 · 5 · 6) = 0.75
∠C = arccos (0.75) ≈ 41.41°
Solution 2. We can use area, via Heron's formula:
Area = √s (s - a)(s - b)(s - c) where s is the semiperimeter (= 15/2)
Area = √(15/2)(3/2)(5/2)(7/2)
A = (15 √7) / 4
Now we can find h: Area = bh/2 = 5h/2 = 15 √7 / 4 → h = 3 √7 / 2
And now, sin A = h/4 = 3 √7 / 8 → ∠A ≈ 82.82°
And sin C = h/6 = 3 √7 / 12 → ∠C ≈ 41.41°.
These solutions are a good advertisement for the law of cosines, don't you think?