We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then
\(θ=\frac{1}{r}\)
Therefore, forr = 100 cm, l = 22 cm, we have
\(θ=\frac{22}{100}\,radian=\frac{180}{\pi}×\frac{22}{100}\,degree=\frac{180×7×22}{22×100}\,degree\)
\(=\frac{126}{10}\,radian=12\frac{3}{5}\,degree=12°36' ........[1°=60']\)
Thus, the required angle is 12°36“²