A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The range (the total distance traveled by the projectile over level ground) of the projectile is approximated by the equation [tex]x = \frac{32}{v^2} sin 2 \theta[/tex]. Find the launch angle of a projectile with an initial speed of 85ft/s and a range of 170 ft.

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation [tex]x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)[/tex], where we will use ft for distances.

From the given equation we have then [tex]\frac{x32ft/s^2}{v^2} =sin(2\theta)[/tex], which means [tex]\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}[/tex] since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have [tex]\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}[/tex], and from now on we have just to use a calculator, obtaining [tex]\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ[/tex]