Question

Given a triangle △ABC with AC=17 , BC=14 and

∠ACB=60 , find the value of AB .

Collected in the board: Law of Cosines

Steven Zheng Steven Zheng posted 1 month ago


Answer

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}

Steven Zheng Steven Zheng posted 1 month ago

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