A displacement vector D is given as $40\mathrm{.}0$ m at an angle of $60\mathrm{.}0$ degrees NORTH of EAST. The $D_x$ component and the $D_y$ component of the vector are

$D_x=+34\mathrm{.}6m,D_y=+20\mathrm{.}0m.$

$D_x=-34\mathrm{.}6m,D_y=+20\mathrm{.}0m.$

$D_x=+20\mathrm{.}0m,D_y=-34\mathrm{.}6m.$

$D_x=+20\mathrm{.}0m,D_y=+34\mathrm{.}6m.$

$D_x=+34\mathrm{.}6m,D_y=-20\mathrm{.}0m.$