The angle of 1"  (second of arc) in radians is  $(Given{360}^0=2\pi rad,1^0=60\text{'}and1\text{'}=60")$

Given, ${360}^0=2\pi rad$

$1^0=\left(\frac{\pi }{180}\right)rad=1.745\times {10}^{-2}rad$

$1^0=60\text{'}=1.745\times {10}^{-2}rad$

$1^{\text{'}}=2.90s\times {10}^{-4}rad\approx 2.91\times {10}^{-4}rad$

$1^{\text{'}}=60"=2.908\times {10}^{-4}rad$

$1"=4.87\times {10}^{-4}rad\approx 4.85\times {10}^{-6}rad$