Answer
$69^\circ$
Work Step by Step
Given the area $A=\frac{1}{2}r^2\theta=15\ cm^2$ and arc length $L=r\theta=6.0\ cm$, we have $r=\frac{6.0}{\theta}$ which can be used in the first relation to give $\frac{1}{2}(\frac{6.0}{\theta})^2\theta=15$. Thus $\theta=\frac{36}{2\times15}=1.2\ rad\approx69^\circ$
