In figure, \(\angle{PQR}\) = \(100^\circ\) ,where P, Q and R are points on a circle with centre O. Find \(\angle{OPR}\).

Since, the angle subtended by the centre is double the angle subtended by circumference, we get,

\(\angle{PQR}\) = 2 \(\angle{PQR}\) = \(200^\circ\)
(given: \(\angle{PQR}\) = \(100^\circ\))

\(\therefore \) in \(\triangle{OPR}\),
\(\angle{POR}\) = \(360^\circ\) - \(200^\circ\) = \(160^\circ\) ...(i)

Again, in \(\triangle{OPR}\),
OP = OR ...(Radii of the circle)
\(\angle{OPR}\) = \(\angle{ORP}\)
(By property of isosceles traingle)

In \(\triangle{POR}\),
\(\angle{OPR}\) + \(\angle{ORP}\) + \(\angle{POR}\) = \(180^\circ\) ...(ii)
(Sum of internal angles of triangle)

\(\therefore \) from eq.(i) and (ii), we get,

\(\angle{OPR}\) + \(\angle{OPR}\) + \(160^\circ\) = \(180^\circ\)
\(\Rightarrow \) 2 \(\angle{OPR}\) = \(180^\circ\) - \(160^\circ\) = \(20^\circ\)
Therefore, \(\angle{OPR}\) = \(10^\circ\).