For this problem, we are given an equation that describes the range of a baseball, and we need to solve it for the angle theta. The equation is shown below:
[tex]r=\frac{1}{32}v^2_o\sin (2\theta)[/tex]
Replacing the variables with the given values, we obtain:
[tex]\begin{gathered} 186=\frac{1}{32}\cdot91^2\cdot\sin (2\theta) \\ 186=\frac{8281}{32}\cdot\sin (2\theta) \\ 5952=8281\cdot\sin (2\theta) \\ \sin (2\theta)=\frac{5952}{8281} \\ \sin (2\theta)=0.7188 \\ \arcsin (\sin (2\theta))=\arcsin (0.7188) \\ 2\theta=45.96º \\ \theta=\frac{45.96º}{2} \\ \theta=22.98º \end{gathered}[/tex]
One possible value for theta is 22.98°.
