Question

in a triangle two sides that measure 6cm and 10cm form an angle that measures 80 degrees. Find to the nearest degree measure of smallest angle of triangle. I found the remaining side was 12 using law of cosines, then tried to use law of sine to find remaining angle but it's not working out, help?

Answer 1

\\[c^2=a^2+b^2-2ab \cos C=6^2+10^2-2(6)(10)\cos (80)\]\[c^2=36+100-120 (0.17)=115.16\]\[c=10.7\]Why is my answer so different than yours? Did I make a mistake?