#### Answer

$x_Q = x_P+d~sin~\theta$
$y_Q = y_P+ d~cos~\theta$

#### Work Step by Step

We can form a right triangle with the points $P$, $Q,$ and $O$, where $O$ is the point $(x_P, y_Q)$
Let $y$ be the side of the triangle adjacent to $\theta$. Let $x$ be the side of the triangle opposite $\theta$. Note that $d$ is the hypotenuse of the triangle.
We can find an expression for $x$:
$\frac{x}{d} = sin~\theta$
$x = d~sin~\theta$
We can find an expression for $y$:
$\frac{y}{d} = cos~\theta$
$y = d~cos~\theta$
We can find an expression for $x_Q$:
$x_Q = x_P+x$
$x_Q = x_P+d~sin~\theta$
We can find an expression for $y_Q$:
$y_Q = y_P+y$
$y_Q = y_P+ d~cos~\theta$