Find the shortest altitude of a triangle with the smallest angle of 30° and two sides containing it of lengths 6cm and 8$\sqrt{3}$cm?

Question

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?

Answer 1

$8 \sqrt 3$ 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 $h=3$