a. Since K is the centre and KR and KS are radii, ∆SKR is isosceles. ∠SRK + ∠RSK + ∠SKR = 180^\circ (sum of angles in a triangle is 180o) ⇒ ∠SRK + ∠RSK + 80^\circ = 180^\circ ⇒ ∠SRK + ∠RSK = 180^\circ - 80^\circ = 100^\circ Since ∠SRK = ∠RSK (base angles of an isosceles triangle are equal) ∠SRK = ∠RSK = \(\frac{100^0}{2} = 50^0\) ∠RSP = 50^\circ + 41^\circ = 91^\circ ∴ ∠ RQP = 180^\circ - 91^\circ = 89^\circ (opp. angles of a cyclic. quad. are supplementary) b. Since KR is a bisector of angle ∠SRQ: ∠SRK = ∠KRQ = 50^\circ ∠SRQ = 50^\circ + 50^\circ = 100^\circ ∴ ∠SPQ = 180^\circ - 100^\circ = 80^\circ (opp. angles of a cyclic. quad. are supplementary)