Solve triangle abc. round decimal to the nearest tenth. a=38, b=31, c=35

Answers: 

Angle A is approximately 70 degrees
Angle B is approximately 50.1 degrees
Angle C is approximately 59.9 degrees

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Work Shown:

Use law of cosines to find angle A
a^2 = b^2 + c^2 - 2*b*c*cos(A)
38^2 = 31^2 + 35^2 - 2*31*35*cos(A)
1444 = 961 + 1225 - 2170*cos(A)
1444 = 2186 - 2170*cos(A)
1444 - 2186  = 2186-2170*cos(A)-2186
-742  = -2170*cos(A)
-2170*cos(A) = -742
-2170*cos(A)/(-2170) = (-742)/(-2170)
cos(A) = 0.341935483870968
arccos(cos(A)) = arccos(0.341935483870968)
A = 70.0051618474206
A = 70 ... round to the nearest tenth
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Use law of cosines to find angle B
b^2 = a^2 + c^2 - 2*a*c*cos(B)
31^2 = 38^2 + 35^2 - 2*38*35*cos(B)
961 = 1444 + 1225 - 2660*cos(B)
961 = 2669 - 2660*cos(B)
961 - 2669  = 2669-2660*cos(B)-2669
-1708  = -2660*cos(B)
-2660*cos(B) = -1708
-2660*cos(B)/(-2660) = (-1708)/(-2660)
cos(B) = 0.642105263157895
arccos(cos(B)) = arccos(0.642105263157895)
B = 50.0510165986799
B = 50.1  ... round to the nearest tenth
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Use law of cosines to find angle C
c^2 = a^2 + b^2 - 2*a*b*cos(C)
35^2 = 38^2 + 31^2 - 2*38*31*cos(C)
1225 = 1444 + 961 - 2356*cos(C)
1225 = 2405 - 2356*cos(C)
1225 - 2405  = 2405-2356*cos(C)-2405
-1180  = -2356*cos(C)
-2356*cos(C) = -1180
-2356*cos(C)/(-2356) = (-1180)/(-2356)
cos(C) = 0.500848896434635
arccos(cos(C)) = arccos(0.500848896434635)
C = 59.9438215538995
C = 59.9  ... round to the nearest tenth
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Or you can use the shortcut
A+B+C = 180
C = 180-A-B
C = 180-70-50.1
C = 59.9
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Check out the attached image for what the triangle looks like. 

Ithink it is b and d