Answer
$\dfrac{\sqrt 3}{2}$
Work Step by Step
The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps:
a) Quadrant- I: $\theta $
b) Quadrant- II: $180^{\circ}-\theta $
c) Quadrant -III: $\theta - 180^o $
d) Quadrant -IV: $360^{\circ}-\theta $
Reference angle of $240^{\circ}$ is equal to
$ =240^{\circ}-180^{\circ}=60^{\circ}$
Since, $\sin 60^{\circ}=\dfrac{\sqrt 3}{2}$
So, $\sin (-240)^{\circ}= \dfrac{\sqrt 3}{2}$; Because $\theta $ lies in Quadrant-II.
