Paper 1 · trigonometry / cosine rule

In triangle $ABC$ , $AB = m$ cm, $AC = 12$ cm and angle $B\hat{A}C = 60°$ . Given that the area of the triangle is $\sqrt{675}$ cm $^2$ , find the exact value of $m$ .

m = 15

m = 5\sqrt{3}

m = 5

m = \sqrt{109}

Explanation:

\tfrac12(12)(m)\sin60°=\sqrt{675}=15\sqrt3

, so

3\sqrt3\,m=15\sqrt3

and

m=5

Derived from ZIMSEC Mathematics Paper 1, November 2012, Q1

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