Question
Given a triangle △ABC with AC=17 , BC=14 and
∠ACB=60 , find the value of AB .
Given a triangle △ABC with AC=17 , BC=14 and
∠ACB=60 , find the value of AB .
Using the Law of Cosines
c^2=a^2+b^2-2ab\cos C
c^2 = 17^2+14^2-2\times 17\times 14\cos 60\degree
c^2 = 17^2+14^2-17\times 14 =17\times 3+14^2
=51+196
=247
\therefore c = \sqrt{247}