Answer
The dog should run 29.6 m at an angle of $71.4^{\circ}$ south of east.
Work Step by Step
We can find the east component $d_x$ of the direction $d$.
$d_x + 12.0~m - 28.0~sin(50.0^{\circ})~m = 0$
$d_x = -12.0~m + 28.0~sin(50.0^{\circ})~m$
$d_x = 9.45~m$
We can find the south component $d_y$ of the direction $d$.
$d_y - 28.0~cos(50.0^{\circ})~m = 10.0~m$
$d_y = 10.0~m + 28.0~cos(50.0^{\circ})~m$
$d_y = 28.00~m$
We can use $d_x$ and $d_y$ to find the magnitude of the distance $d$.
$d = \sqrt{(d_x)^2+(d_y)^2}$
$d = \sqrt{(9.45~m)^2+(28.00~m)^2}$
$d = 29.6~m$
We can find the angle south of east.
$tan(\theta) = \frac{28.00}{9.45}$
$\theta = tan^{-1}(\frac{28.00}{9.45}) = 71.4^{\circ}$
The dog should run 29.6 m at an angle of $71.4^{\circ}$ south of east.
