3. 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: \(\angle{PQR}\) = \(100^\circ\))

So, 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)

So, from eq.(i) and (ii), we get,

\(\angle{OPR}\) + \(\angle{OPR}\) + \(160^\circ\) = \(180^\circ\)

i.e., 2 \(\angle{OPR}\) = \(180^\circ\) - \(160^\circ\) = \(20^\circ\)

Therefore, \(\angle{OPR}\) = \(10^\circ\).