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

Solution:

(i) ∠AOC = 135°

∴ ∠AOC + ∠BOC = 180° [Linear pair of angles]

⇒ 135° + ∠BOC = 180°

⇒ ∠BOC = 180° − 135°

⇒ ∠BOC = 45°

By degree measure theorem

∠BOC = 2∠CPB

⇒ 45° = 2x

⇒ x = 45°/2 = 22½°

(ii) We have ∠ABC = 40°

∠ACB = 90° [Angle in semi circle]

In ΔABC, by angle sum property

∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 90° + 40° = 180°

⇒ ∠CAB = 180° − 90° − 40°

⇒ ∠CAB = 50°

Now, ∠CDB = ∠CAB [Angle is same in segment]

⇒ x = 50°

(iii) We have

∠AOC = 120° By degree measure theorem.

∠AOC = 2∠APC

⇒ 120° = 2∠APC

⇒ ∠APC = 120°/2 = 60°

∠APC + ∠ABC = 180° [Opposite angles of cyclic quadrilaterals]

⇒ 60° + ∠ABC = 180°

⇒ ∠ABC = 180° − 60°

⇒ ∠ABC = 120°

∴ ∠ABC + ∠DBC = 180° [Linear pair of angles]

⇒ 120 + x = 180°

⇒ x = 180° − 120° = 60°

(iv)  We have

∠CBD = 65°

∴ ∠ABC + ∠CBD = 180° [Linear pair of angles]

⇒ ∠ABC = 65° = 180°

⇒ ∠ABC = 180° − 65° = 115°

∴ reflex ∠AOC = 2∠ABC [By degree measure theorem]

⇒ x = 2 × 115°

⇒ x = 230°

(v) We have

∠OAB = 35° Then,

∠OBA = ∠OAB = 35° [Angles opposite to equal radii]

In ΔAOB, by angle sum property

⇒∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 35° + 35° = 180°

⇒ ∠AOB = 180° −35° − 35° = 110°

∴ ∠AOB + reflex ∠AOB = 360° [Complexangle]

⇒ 110° + reflex ∠AOB = 360°

⇒ reflex ∠AOB = 360° − 110° = 250°

By degree measure theorem reflex

∠AOB = 2∠ACB

⇒ 250° = 2x

⇒ x = 250°/2 = 125°

(vi) We have

∠AOB = 60° By degree measure theorem reflex

∠AOB = 2∠ACB

⇒ 60° = 2∠ACB

⇒ ∠ACB = 60°/2 = 30° [Angles opposite to equal radii]

⇒ x = 30°.

(vii) We have

∠BAC = 50° and ∠DBC = 70°

∴ ∠BDC = ∠BAC = 50° [Angle in same segment]

In ΔBDC, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180°

⇒ 50° + x + 70° = 180°

⇒ x = 180° − 50° − 70° = 60°

(viii) We have

∠DBO = 40° and ∠DBC = 90° [Angle in a semicircle]

⇒ ∠DBO + ∠OBC = 90°

⇒ 40° + ∠OBC = 90°

⇒ ∠OBC = 90° − 40° = 50° By degree measure theorem

∠AOC = 2∠OBC

⇒ x = 2 × 50° = 100°

(ix)  In ΔDAB, by angle sum property

∠ADB + ∠DAB + ∠ABD = 180°

⇒ 32° + ∠DAB + 50° = 180°

⇒ ∠DAB = 180° − 32° − 50°

⇒ ∠DAB = 98°

Now, ∠OAB + ∠DCB = 180° [Opposite angles of cyclic quadrilateral]

⇒ 98° + x = 180°

⇒ x = 180° − 98° = 82°

(x) We have,

∠BAC = 35°

∠BDC = ∠BAC = 35° [Angle in same segment]

In ΔBCD, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180°

⇒ 35° + x + 65° = 180°

⇒ x = 180° − 35° − 65° = 80°

(xi) We have

∠ABD = 40°

∠ACD = ∠ABD = 40° [Angle in same segment]

In ΔPCD, by angle sum property

∠PCD + ∠CPO + ∠PDC = 180°

⇒ 40° + 110° + x = 180°

⇒ x = 180° − 150°

⇒ x = 30°

(xii) Given that,

∠BAC = 52°

Then ∠BDC = ∠BAC = 52° [Angle in same segment]

Since OD = OC

Then ∠ODC = ∠OCD [Opposite angle to equal radii]

⇒ x = 52°