Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Find the answer

Find the area of triangle $ABC$ given that $AB=8$ , $AC=3$ , and $\angle BAC=60^{\circ}$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Using the law of cosines gives: $\begin{aligned} x^{2}+(11-x)^{2}-2x(11-x) \cos 60^{\circ} & =7^{2} \\ 3x^{2}-33x+72 & =0 \\ x & =3 \text{ or } 8. \end{aligned}$ Therefore, $AB=8$ and $AC=3$ or $AB=3$ and $AC=8$ . In both cases, the area of the triangle is: $\frac{1}{2} \cdot 8 \cdot 3 \sin 60^{\circ}=6 \sqrt{3}$ .

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