Pre calculus please help!

I use the law of sines and cosines to do these but i keep getting the wrong answers.

1) If a=4, b=9 and c=10, find the angles A, B and C.

2)Two ships leave a harbor at the same time, traveling on courses that have an angle of 140 degrees between them. If the first ship travels at 30 miles per hour and the second ship travels at 31 miles per hour, how far apart are the two ships after 2.8 hours?

3) a triangle ABC with sides ac=10 bc= 18 and angle C =120 degrees.
find the measure of side c, angle A and angle B


thank you for your help!

1) a^2 = b^2 + c^2 - 2bcCosA
16 = 81 + 100 - 2*9*10CosA
180CosA = 165
Cos A = 0.9167
Angle A = 23.56°
Repeating for Angle B:
b^2 = a^2 + c^2 - 2acCos B
Cos B = 35/80
Angle B = 64.06°

Angle C = 180 -A - B = 92.38°


2) Your triangle has one side = 30x2.8 = 84 miles, 2nd side is 86.8 miles and included angle is 140°
x^2 = 84^2 + 86.8^2 - 2 x 84 x 86.8 Cos 140
= 7056 + 7534 - (14582 x -0.766)
= 14590 + 11170 = 25752
Distance = 160.5 miles


3) c^2 = 10^2 + 18^2 - 2x10 x18 Cos 120
= 100 + 324 + 180
= 604
c = 24.58

c/Sinc C = b/Sin B
24.58 / Sin 120 = 10 / Sin B
Sin B = 10 / 28.38 = 0.3524
Angle B = 20.63°

Angle A = 180 -120 - 20.63 = 39.37°