Answer
$\dfrac{1}{2}$
Work Step by Step
$270 < A < 360$
Dividing by 2
$135 < \dfrac{A}{2} < 180$
$\therefore \dfrac{A}{2} $ terminates in $QII$
$\sin{\dfrac{A}{2}} = \sqrt{\dfrac{1-\cos{A}}{2}}= \sqrt{\dfrac{1-0.5}{2}}$
$\sin{\dfrac{A}{2}} = \boxed{\dfrac{1}{2}}$
