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, $$\angle{OPR}$$ = $$10^\circ$$.