(i) A circle with centre O
∠AOC = 135∘
But ∠AOC + ∠COB = 180∘ (Linear pair)
⇒ 135∘ + ∠COB = 180∘
⇒ ∠COB = 180∘ − 135∘ = 45∘
Now are BC subtends ∠BOC at the centre and ∠BPC at the remaining part of the circle
∴ ∠BOC = 2∠BPC
⇒ ∠BPC = 1/2 ∠BOC
= 12 × 45∘
= \(\frac{45°}2\)
∴ x = 22 1/2∘
(ii) ∵ CD and AB are the diameters of the circle with centre O
∠ABC = 40∘ But in ΔOBC,
OB = OC (Radii of the circle)
∴ ∠OCB = ∠OBC = 40∘
Now in ΔBCD,
∠ODB + ∠OCB + ∠CBD = 180∘
⇒ x + 40∘ + 90∘ = 180∘ (Angles of a triangle)
⇒ x + 130∘ = 180∘
⇒ x = 180∘ − 130∘ = 50∘
(iii) In circle with centre O, ∠AOC = 120∘, AB is produced to D
∵∠AOC = 120∘
and ∠AOC + convex ∠AOC = 360∘
⇒120∘ + convex ∠AOC = 360∘
∴ Convex ∠AOC = 360∘ − 120∘ = 240∘
∴ are APC Subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle
∴ ∠ABC = 1/2 ∠AOC = 1/2 × 240∘ = 120∘
But ∠ABC at the remaining part of the circle
∴ ∠ABC = 1/2 ∠AOC = 1/2 × 240∘ = 120∘
But ∠ABC + ∠CBD = 180∘(Linear Pair)
⇒ 120∘ + x = 180∘
⇒ x = 180∘ − 120∘ = 60∘
∴ x = 60∘
(iv) A circle with centre O and ∠CBD = 65∘
But ∠ABC + ∠CBD = 180∘(Linear Pair)
⇒ ∠ABC + 65∘= 180∘
⇒ ∠ABC = 180∘ − 65∘ = 115∘
Now are AEC subtends ∠x at the centre and ∠ABC at the remaining part of the cirlce
∴ ∠AOC = 2∠ABC
⇒ x = 2 × 115∘= 230∘
∴ x = 230∘
(v) In circle with centre O AB is chord of the cirlce, ∠OAB = 35∘
In △OAB,
OA = OB
∴ OBA = ∠OAB = 35∘
But in △OAB
∠OAB + ∠OBA + ∠AOB = 180∘
⇒ 35∘ + 35∘ + ∠AOB = 180∘
⇒ 70∘ + ∠AOB = 180∘
⇒ ∠AOB = 180∘ − 70∘ = 110∘
∴Convex ∠AOB = 360∘ − 10∘ = 250∘
But are AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠ACB = 1/52 ∠AOB
⇒ x = 12 × 250∘ = 125∘
⇒ x = 125∘
(vi) In the circle with centre O, BOC is its diameter, ∠AOB = 60∘
Are AB subtends ∠AOB at the centre of the circle and ∠ACB at the remaining part of he circle
∴ ∠ACB = 1/2∠AOB
= 1/2 × 60∘
= 30∘
But in △OAC,
OC = OA
∴ ∠OAC = ∠OCA = ∠ACB
⇒ x = 30∘
(vii) In the circle, ∠BAC and ∠BDC are in the same segment
∠DBC + ∠BCD + ∠BDC = 180∘
⇒ 70∘ + x + 50∘ = 180∘
⇒ x + 120∘ = 180∘
⇒ x = 180∘ − 120∘ = 60∘
∴ x = 60∘
(viii) In circle with centre O,
∠OBD = 40∘
AB and CD are diameters of the circle
∠DBA and ∠ACD are in the same segment
∴ ∠ACD = ∠DBA = 40∘
In △OAC,
OA = OC
∴ ∠OAC = ∠OCA = 40∘
and ∠OAC + ∠OCA + ∠AOC = 180∘
⇒ 40∘ + 40∘ + x = 180∘
⇒ x + 80∘ = 180∘
⇒ x = 180∘ − 80∘ = 100∘
∴ x = 100∘
(ix) In the circle, ABCD is a cyclic quadrilateral ∠ADB = 32∘, ∠DAC = 28∘ and ∠ABD = 50∘
∠ABD and ∠ACD are in the same segment of a circle
∴ ∠ABD = ∠ACD
⇒ ∠ACD = 50∘
Similarly, ∠ADB = ∠ACB
⇒ ∠ACB = 32∘
Now ∠DCB = ∠ACD + ∠ACB
= 50∘ + 32∘ = 82∘
∴ x = 82∘
(x) In a circle,
∠BAC = 35∘, ∠CBD = 65circ
∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 35∘
In △BCD,
∠BDC + ∠BCD + ∠CBD = 180∘
⇒ 35∘ + x + 65∘ = 180∘
⇒ x + 100 = 180∘
⇒ x + 100∘ = 180∘
⇒ x − 180∘ − 100∘ = 80∘
∴ x = 80∘
(xi) In the circle,
∠ABD and ∠ACD are in the same segment of a circle
∴ ABD = ∠ACD = 40∘
Now in △CPD,
∠CPD + ∠PCD + ∠PDC = 180∘
⇒ 110∘ + 40∘ + x = 180∘
⇒ x + 150∘ = 180∘
∴ x = 180∘ − 150∘ = 30∘
(xii) In the circle, two diameters AC and BD intersect each other at O
∠BAC = 50∘
OA = OB
∴ ∠OBA = ∠OAB = 52∘
⇒ ∠ABD = 52∘
But ∠ABD and ∠ACD are n the same segment of the circle
∴ ∠ABD = ∠ACD
⇒ 52∘ = x
∴ x = 52∘