# Match the following columns: Column I Column II (a) Angle in a semicircle measures (p) 40° (b) In the given figure, O is the centre of a circle. If ∠AOB = 120°, then ∠ACB = ? (q) 80° (c) In the given figure, O is the centre of a circle. If ∠POR = 90° and ∠POQ = 110°, then ∠QPR = ? (r) 90° (d) In cyclic quadrilateral ABCD, it is given that ∠ADC = 130° and AOB is a diameter of the circle through A, B, C and D. Then, ∠BAC = ? (s) 60° (a) ....., (b) ....., (c) ....., (d) .....,

(a) Angle in a semicircle measures 90°.
(b) Now, chord AB subtends ∠AOB at the centre and ∠ACB at a point C of the remaining part of a circle.

(c) ∠QOR = {360° - (110° + 90° )} = (360° - 200°) = 160°

∠QPR = $\frac{1}{2}\angle \mathrm{QOR}=\left(\frac{1}{2}×160°\right)=80°$
(d) Since ∠ADC + ∠ABC = 180°  (Opposite angles in a cyclic quadrilateral)

⇒ 130° + ∠ABC = 180°

∠ABC = (180° - 130°) = 50°
and ∠ACB = 90°    (Angle in a semicircle)
In
ΔABC, we have:
∠ABC + ∠ACB + ∠BAC = 180°   (Angle sum property of a triangle)

⇒ 50° + 90° + ∠BAC = 180°

∠BAC = (180° - 140°) = 40°

Hence, (a) - (r), (b) - (s), (c) - (q) and (d) - (p)

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