By definition of the trig functions,
[tex]\sin30^\circ=\dfrac y{16}[/tex]
[tex]\cos30^\circ=\dfrac x{16}[/tex]
We have [tex]\sin30^\circ=\dfrac12[/tex], so [tex]y=8[/tex], and [tex]\cos30^\circ=\dfrac{\sqrt3}2[/tex], so [tex]x=8\sqrt3[/tex].
Just to check: we should have
[tex]x^2+y^2=16^2[/tex]
by the Pythagorean theorem. Indeed,
[tex]8^2+(8\sqrt3)^2=4\cdot8^2=2^2(2^3)^2=2^8=(2^4)^2=16^2[/tex]
