Solve for θ :
r = 1/32 v₀² sin(2θ)
32r / v₀² = sin(2θ)
arcsin(32r / v₀²) = 2θ
⇒ θ = 1/2 arcsin(32r / v₀²)
Note that in projectile motion, we're only concerned with launch angles in the range 0 ≤ θ ≤ π/2, so there are no issues with taking inverse sines.
