Answer :

(a) Angle in a semicircle measures – 90° (r)


(b) In the given figure, O is the centre of a circle. If AOB = 120°, then ACB = ?



AOB = ACB


ACB = × 120° = 60°


ACB = 60° (s)


(c) In the given figure, O is the centre of a circle. If and then



Here, OP = OR = OQ (radius)


In ΔPOR


OPR = ORP (angles opposite to equal sides are equal)


By angle sum property


POR + OPR + ORP = 180°


90° + 2×OPR = 180°


OPR = 180°—90°


OPR = 90°


OPR = 45°


Similarly in Δ POQ


OPQ = OQP (angles opposite to equal sides are equal)


By angle sum property


POQ + OPQ + OQP = 180°


110° + 2×OQP = 180°


OQP = 180°–110°


OQP = 70°


OQP = 35°


QPR = QPO + OPR = 45° + 35° = 80°


QPR = 80° (q)


(d) In cyclic quadrilateral it is given that and is a diameter of the circle


through and Then,



Here,


ADC + ABC = 180° (opposite angles in cyclic quadrilateral are supplymentary)


130° + ABC = 180°


ABC = 180°–130° = 50°


In ΔABC


By angle sum property


BAC + ABC + ACB = 180°


BAC + 50° + 90° = 180°


BAC = 180°–50°–90° = 40°


∴∠BAC = 40° (p)


Answers are: (a) – (r), (b) – (s), (c) – (q), (d) – (p)


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