Question

How do you solve this

Answer 1

By definition of the trig functions,

$\sin30^\circ=\dfrac y{16}$

$\cos30^\circ=\dfrac x{16}$

We have $\sin30^\circ=\dfrac12$, so $y=8$, and $\cos30^\circ=\dfrac{\sqrt3}2$, so $x=8\sqrt3$.

Just to check: we should have

$x^2+y^2=16^2$

by the Pythagorean theorem. Indeed,

$8^2+(8\sqrt3)^2=4\cdot8^2=2^2(2^3)^2=2^8=(2^4)^2=16^2$