Question

How do you solve this

Answer 1

By definition of the trig functions,

`\sin30^\circ=\frac y{16}`

`\cos30^\circ=\frac x{16}`

We have
`\sin30^\circ=\frac12`, so
`y=8`, and
`\cos30^\circ=\frac{\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`