Answer
$$A= 80^{\circ}+n360^{\circ},\ \ \text{or}\ \ 20^{\circ}+n360^{\circ} $$
where $n $ is integer
Work Step by Step
Given
$$ \cos (A-50^{\circ}) =\left( \frac{\sqrt{3}}{2}\right)$$
Since
\begin{align*}
A-50^{\circ}&=\cos^{-1}\left( \frac{\sqrt{3}}{2}\right)\\
A-50^{\circ}&=30^{\circ},\ \ \text{or}\ \ 330^{\circ}\\
A&=80^{\circ},\ \ \text{or}\ \ 20^{\circ}
\end{align*}
Since the period of $\cos$ function is $360^{\circ}$ , then the general solution is
$$A= 80^{\circ}+n360^{\circ},\ \ \text{or}\ \ 20^{\circ}+n360^{\circ} $$
where $n $ is integer