Find (to the nearest degree) the largest angle of a triangle whose sides measure 5 inches, 8 inches, and 12 in

I got an answer of 24.6 but I'm not entirely sure I did it correctly.

since 12 is the largest side, opposite of it is the largest angle
thus,
let c = 12 in; C is the largest angle
a & b are the other sides which measures 5 and 8 in respectively

c² = a² + b² - 2abcosC
12² = 5² + 8² - 2(5)(8)cosC
12² - 5² - 8² = -80cosC
55 = -80cosC
-55/80 = cos C
C = arccos -11/16
C = 133.4325°
C = 133.43°

Largest angle will be opposite largest side: 12

Use law of cosines:
c² = a² + b² − 2ab cos(C)
2ab cos(C) = a² + b² − c²
cos(C) = (a² + b² − c²) / (2ab)

cos(C) = (5² + 8² - 12²) / (2*5*8)
cos(C) = −55/80 = −11/16
C = arccos(−11/16)
C = 133.43°