Answer
$225^{\circ}$
Work Step by Step
Given-
$\sin\theta$ = - $\frac{\sqrt 2}{2}$ , $\theta$ belongs to Q III
We will find reference angle of $\theta$ first using positive value $\frac{\sqrt 2}{2}$ and reference angle theorem-
$\sin\theta$ = - $\frac{\sqrt 2}{2}$ = -$\sin 45^{\circ}$ (As $\sin 45^{\circ}$ = $\frac{\sqrt 2}{2}$ )
Therefore by reference angle theorem, $45^{\circ}$ is the reference angle for desired angle $\theta$
The desired angle $\theta$ is in Q III,
Also $0^{\circ}\leq \theta\lt 360^{\circ}$
Therefore-
$\theta$ = $180^{\circ} + 45^{\circ}$
= $225^{\circ}$
