Answer
$$A= 70^{\circ}+n360^{\circ},\ \ \text{or}\ \ 10^{\circ}+n360^{\circ} $$
where $n $ is integer
Work Step by Step
Given
$$ \sin (A+50^{\circ}) =\left( \frac{\sqrt{3}}{2}\right)$$
Since
\begin{align*}
A+50^{\circ}&=\sin^{-1}\left( \frac{\sqrt{3}}{2}\right)\\
A+50^{\circ}&=60^{\circ},\ \ \text{or}\ \ 120^{\circ}\\
A&=10^{\circ},\ \ \text{or}\ \ 70^{\circ}
\end{align*}
Since the period of $\sin$ function is $360^{\circ}$ , then the general solution is
$$A= 70^{\circ}+n360^{\circ},\ \ \text{or}\ \ 10^{\circ}+n360^{\circ} $$
where $n $ is integer