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°

2×∠OPR = 180°—90°

2×∠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°

2×∠OQP = 180°–110°

2×∠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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