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° |
(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 =
(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)
(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 =
(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)