Approximating $\frac{\pi}{2}$ from above
Since $$\left(\frac{\pi}{2}\right)^9\approx 58.220897$$ the root $$58^\frac{1}{9}\approx 1.5701$$ is not far from $$\frac{\pi}{2}\approx 1.570796$$
This approximation may be improved by noticing that $$0.22\approx \frac{2}{9} = 0.22...\,$$
so $$\frac{\pi}{2} \approx \left(58+\frac{2}{9}\right)^\frac{1}{9}\approx 1.570800$$
Approximating $\frac{\pi}{2}$ from below
A similar approximation is given by $$\left(1+\frac{2}{9}\right)^\frac{9}{4} \approx 1.5707$$ which may be extended to
$$\frac{\pi}{2} \approx \left(1+\frac{2}{9}+\frac{1}{2^3·5^5}\right)^\frac{9}{4} \approx 1.5707963$$ to yield seven correct decimals.
Alternatively,
$$\frac{\pi}{2} \approx \left(58+\frac{2}{9}\right)^\frac{1}{9}-\frac{1}{2^4·5^6}\left(1-\frac{13}{3·5^4}\right) \approx 1.57079632679437$$
has twelve correct decimals.
Inequalities for $\frac{\pi}{2}$ and $\log(2)$
Combining both approximations we may write $$\left(58+\frac{2}{9}\right)^\frac{1}{9}-\frac{1}{2^4·5^6}\left(1-\frac{13}{3·5^4}\right)<\frac{\pi}{2}<\left(58+\frac{2}{9}\right)^\frac{1}{9}$$
following the pattern of the double inequality for $\log(2)$ $$\left(\frac{2}{5}\right)^\frac{2}{5}<\log(2)<\left(\frac{1}{3}\right)^\frac{1}{3}$$
taken from a comment to this question.
Q Is this 58 related to the almost-integer $e^{\sqrt{58}\pi} \approx24591257751.99999982$ ?
(see equation (68) here)
