Question

If O is the centre of the circle, find the value of x in each of the following figures :

Answer 1

(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 = 65^circ 

∠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^∘

Answer 2

(i) ∠AOC = 135° 

Therefore, 

∠AOC + ∠BOC = 180° 

(Linear pair) 

135° + ∠BOC = 180° 

∠BOC = 45° 

By degree measure theorem, 

∠BOC = 2∠COB 

45° = 2x 

x = 22\(\frac{1}{2}\)°  

(ii) We have, 

∠ABC = 40° 

∠ACB = 90° 

(Angle in semi-circle) 

In triangle ABC, 

By angle sum property,

∠CAB + ∠ACB + ∠ABC = 180° 

∠CAB + 90° + 40° = 180° 

∠CAB = 50° 

Now, 

∠COB = ∠CAB 

(Angle on same segment) 

x = 50° 

(iii) We have,

∠AOC = 120° 

By degree measure theorem, 

∠AOC = 2∠APC 

120°= 2∠APC 

∠APC = 60° 

Therefore, 

∠APC + ∠ABC = 180° 

(Opposite angles of cyclic quadrilateral) 

60° + ∠ABC = 180° 

∠ABC = 120° 

∠ABC + ∠DBC = 180°

(Linear pair) 

120° + x = 180° 

x = 60° 

(iv) We have, 

∠CBD = 65° 

Therefore, 

∠ABC + ∠CBD = 180° 

(Linear pair) 

∠ABC + 65° = 180° 

∠ABC = 115° 

Therefore, 

Reflex ∠AOC = 2∠ABC 

(By degree measure theorem) 

x = 2 x 115° = 230° 

(v) We have,

∠OAB = 35°

Then,

∠OBA = ∠OAB = 35° 

(Angle opposite to equal sides are equal) 

In triangle AOB, 

By angle sum property, 

∠AOB + ∠OAB + ∠OBA = 180° 

∠AOB + 35° + 35° =180°

∠AOB = 110° 

Therefore, 

∠AOB + Reflex ∠AOB = 360° 

(Complete angle) 

110° + Reflex ∠AOB = 360° 

Reflex ∠AOB = 250° 

By degree measure theorem, 

Reflex ∠AOB = 2∠ACB 

250° = 2x 

x = 125° 

(vi) We have, 

∠AOB = 60° 

By degree measure theorem, 

∠AOB = 2∠ACB 

60° = 2∠ACB 

∠ACB = 30° x = 30° 

(vii) We have, 

∠BAC = 50° 

∠DBC = 70° 

Therefore, 

∠BDC = ∠BAC = 50° 

(Angles on same segment) 

In triangle BDC, 

By angle sum property 

∠BDC + ∠BCD + ∠DBC = 180° 

50° + x + 70° = 180° 

120° + x = 180° 

x = 60° 

(viii) We have, 

∠DBO = 40° 

∠DBC = 90° 

(Angle in semi-circle) 

Therefore, 

∠DBO + ∠OBC = 90° 

40° + ∠OBC = 90° 

∠OBC = 50° 

By degree measure theorem, 

∠AOC = 2∠OBC x = 2 x 50° 

x = 100° 

(ix) In triangle DAB, 

By angle sum property 

∠ADB + ∠DAB + ∠ABD = 180° 

32° + ∠DAB + 50°= 180° 

∠DAB = 98° 

Now, 

∠DAB + ∠DCB = 180° 

(Opposite angle of cyclic quadrilateral) 

98° + x = 180° 

x = 180° – 98° = 82° 

(x) We have, 

∠BAC = 35° 

∠BDC = ∠BAC = 35° 

(Angle on same segment) 

In triangle BCD, 

By angle sum property 

∠BDC + ∠BCD + ∠DBC = 180° 

30° + x + 65° = 180° 

x = 80° 

(xi) We have, 

∠ABD = 40° 

∠ACD = ∠ABD = 40° 

(Angle on same segment) 

In triangle PCD, 

By angle sum property 

∠PCD + ∠CPD + ∠PDC = 180° 

40° + 110° + x = 180° 

x = 30° 

(xii) Given that, 

∠BAC = 52° 

∠BDC = ∠BAC = 52° 

(Angle on same segment) 

Since, 

OD = OC 

Then, 

∠ODC = ∠OCD 

(Opposite angles to equal radii) 

x = 52°