Solve triangle ABC, when A = 6, B=10 and c=12

Use cosine rule,

cos (A) = (b^2+c^2-a^2) / (2bc)

= (10^2+12^2-6^2) / (2*10*12)

=13/15

A=29.926 degrees ... (A)

cos (B) = (c^2+a^2-b^2) / (2ca)

= (12^2+6^2-10^2) / (2*12*6)

=5/9

B=56.251 degrees ... (B)

cos (C) = (a^2+b^2-c^2) / (2ab)

= (6^2+10^2-12^2) / (2*6*10)

=-1/15

C=93.823 degrees ... (C)

Check:29.926+56.251+93.823=180.0 degrees ... ok