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I'm not really sure how to approach this problem, but since I have two sides and an angle maybe I could use area=$\frac{absin(C)}{2}$. What would be the best way to approach this problem?

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$8 \sqrt 3$ must be the longest side, so the shortest altitude would be the height corresponding to that side

so $$ \frac 12 (8\sqrt 3) h = \frac{6(8\sqrt 3)\sin(30)}{2} $$

which gives $h=3$

WW1
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