[tex]cos(30^o)= \frac{adjacent}{hypotenuse} = \frac{x}{16 \sqrt{3} } \ \ \to \\x=16 \sqrt{3}*cos(30^o) =16 \sqrt{3}* \cfrac{ \sqrt{3} }{2} =8*3=24 \\ \\ \\ sin(30^o)= \frac{opposite }{hypotenuse} =\frac{y}{16 \sqrt{3} } \ \ \to \\y=16 \sqrt{3}*sin(30^o) =16 \sqrt{3}* \cfrac{1 }{2} =8 \sqrt{3} [/tex]
Answer: C.
