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You know that Area = base x height / 2, but what other ways are there to find the area of a triangle? Brace yourself for some potent formulae.

In triangle \(ABC,\) \(AB=3,\) \(AC = 12,\) and \(\sin \angle BAC = 0.5.\) What is the area of triangle \(ABC?\)

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In \(\triangle ABC,\) \(AC = 3\) and \(\sin \angle ACB = \frac{2}{3}.\) If the area of \(\triangle ABC\) is \(9,\) what is \(BC?\)

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In triangle \(ABC\) above, \(\overline{AD}\) bisects \(\overline{BC}.\) Given that \(AB = 12,\) \(AD =8,\) and \(\sin \angle BAD = \frac{1}{4},\) what is the area of \(\triangle ABC?\)

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In triangle \(ABC,\) \(\overline{AD}\) bisects \(\angle BAC,\) \(\overline{AE}\) is perpendicular to \(\overline{BC},\) and \(F\) and \(G\) are the midpoints of \(\overline{AB}\) and \(\overline{AC},\) respectively. Given that \(AB = 16,\) \(AC=13,\) \(AH=13,\) and \(AI=8,\) and \(\angle BAC = 90^{\circ},\) which of the following triangles has an area equal to half the area of \(\triangle ABC?\)

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In triangle \(ABC,\) \(AB = 3\) and \(BC=6.\) If the area of the triangle is \(6,\) what is \(\sin \angle ABC?\)

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