Angle & Cyclic Properties of a Circle

Join AC.

As, OA = OC = radius of circle.

Let ∠OAC = ∠OCA = x (As angles opposite to equal sides are equal)

We know that,

Sum of angles of triangle = 180°

∴ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° - 2x

From figure,

⇒ ∠BAC = ∠BAO + OAC = 30° + x

⇒ ∠BCA = ∠BCO + OCA = 40° + x

Now, in ∆ABC

⇒ ∠ABC = 180° - ∠BAC - ∠BCA [Angle sum property of a triangle]

= 180° - (30° + x) - (40° + x)

= 180° - 30° - x - 40° - x

= 180° - 70° - 2x

= 110° - 2x.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOC = 2∠ABC

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ -2x + 4x = 220° - 180°

⇒ 2x = 40°

⇒ x = 20°.

Thus, ∠AOC = 180° - 2x = 180° - 2(20°)

= 180° - 40° = 140°.

Hence, ∠AOC = 140°.

From figure,

⇒ ∠AOC + Reflex ∠AOC = 360°

⇒ 130° + Reflex ∠AOC = 360°

⇒ Reflex ∠AOC = 360° - 130°

⇒ Reflex ∠AOC = 230°.

Arc AC subtends Reflex ∠AOC at center and ∠ABC at another point of circle.

⇒ Reflex ∠AOC = 2 ∠ABC

⇒ 2∠ABC = 230°

⇒ ∠ABC = $\frac{230^{\circ}}{2}$

⇒ ∠ABC = 115°.

Hence, the value of ∠ABC = 115°.

(i) We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠AOB = 2∠ACO

⇒ 110° = 2∠ACO

⇒ ∠ACO = $\frac{110^{\circ}}{2}$

⇒ ∠ACO = 55°

Hence, ∠ACO = 55°.

(ii) From figure,

⇒ ∠COA + ∠AOB = 180° [Linear pair]

⇒ ∠COA + 110° = 180°

⇒ ∠COA = 180° - 110°

⇒ ∠COA = 70°

The sum of the three interior angles of any triangle is always 180°.

In ΔAOC,

⇒ ∠COA + ∠ACO + ∠CAO = 180°

⇒ 70° + 55° + ∠CAO = 180°

⇒ 125° + ∠CAO = 180°

⇒ ∠CAO = 180° - 125°

⇒ ∠CAO = 55°.

Hence, ∠CAO = 55°.

(i) We know that:

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠BAD + ∠BCD = 180°

⇒ ∠BCD = 180° - 100°

⇒ ∠BCD = 80°.

Hence, ∠BCD = 80°.

(ii) Since AB ∥ DC, the angles ∠BAD and ∠ADC are consecutive interior angles along the transversal AD.

Therefore,

⇒ ∠BAD + ∠ADC = 180°

⇒ 100° + ∠ADC = 180°

⇒ ∠ADC = 180° - 100°

⇒ ∠ADC = 80°.

Hence, ∠ADC = 80°.

(iii) We know that:

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 80° = 180°

⇒ ∠ABC = 180° - 80°

⇒ ∠ABC = 100°.

Hence, ∠ABC = 100°.

(i) From figure,

∠ADB = ∠ACB = 52° [Angles in the same segment are equal]

∠ADB = 52°.

Hence, ∠ADB = 52°.

(ii) From figure,

∠BAC = ∠BDC = 43° [Angles in the same segment are equal]

∠BAC = 43°.

Hence, ∠BAC = 43°.

(iii) We know that,

The sum of the three interior angles of any triangle is always 180°.

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 43° + ∠ABC + 52° = 180°

⇒ ∠ABC + 95° = 180°

⇒ ∠ABC = 180° - 95°

⇒ ∠ABC = 85°.

Hence, ∠ABC = 85°.

(i) From figure,

OA = OC [Radius of same circle]

In Δ AOC,

∠OCA = ∠OAC = 50° (As angles opposite to equal sides in a triangle are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

∠AOC + ∠OCA + ∠OAC = 180°

∠AOC + 50° + 50° = 180°

∠AOC = 180° − 50° − 50° = 80°.

We know that,

Angle at the center is double the angle at the circumference subtended by the same chord.

∠ABC = $\frac{1}{2}$ ∠AOC

∠ABC = $\frac{1}{2}$ 80°

∠ABC = 40°.

Hence, ∠ABC = 40°.

(ii) From figure,

∠BOC = ∠AOB − ∠AOC = 140° − 80° = 60°.

OB = OC (Radius of same circle)

Let ∠OBC = ∠OCB = x (As angles opposite to equal sides are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\frac{120^{\circ}}{2}$

⇒ x = 60°.

Hence, ∠BCO = 60°.

(iii) In ∆AOB, we have

OA = OB (radius of same circle)

So, ∠OBA = ∠OAB (As angles opposite to equal sides are equal)

By angle sum property of a triangle we get,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 140° = 180°

⇒ 2∠OAB = 40°

⇒ ∠OAB = $\frac{40^{\circ}}{2}$

⇒ ∠OAB = 20°.

Hence, ∠OAB = 20°.

(iv) We know that,

⇒ ∠BCO = 60°

⇒ ∠OCA = 50°

From figure,

⇒ ∠BCA = ∠BCO + ∠OCA

⇒ ∠BCA = 60° + 50°

⇒ ∠BCA = 110°.

Hence, ∠BCA = 110°.

(i) By angle sum property of a triangle we get,

In ΔABD,

⇒ ∠BAD + ∠ABD + ∠ADB = 180°

⇒ 70° + 50° + ∠ADB = 180°

⇒ 120° + ∠ADB = 180°

⇒ ∠ADB = 180° - 120°

⇒ ∠ADB = 60°.

From figure,

⇒ ∠BDC = ∠ADC - ∠ADB

⇒ ∠BDC = 80° - 60°

⇒ ∠BDC = 20°.

Hence, ∠BDC = 20°.

(ii) We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠BAD + ∠BCD = 180°

⇒ 70° + ∠BCD = 180°

⇒ ∠BCD = 180° - 70°

⇒ ∠BCD = 110°.

Hence, ∠BCD = 110°.

(iii) We know that,

Angles in the same segment of a circle are equal.

⇒ ∠BCA = ∠ADB

⇒ ∠BCA = 60°.

Hence, ∠BCA = 60°.

From figure,

∠ACB = 90° [Angle in a semicircle is a right angle]

We know that,

In a cyclic quadrilateral, the sum of opposite angles is 180°.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140°

⇒ ∠ABC = 40°.

By angle sum property of a triangle we get,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ ∠BAC + 40° + 90° = 180°

⇒ ∠BAC + 130° = 180°

⇒ ∠BAC = 180° - 130°

⇒ ∠BAC = 50°.

Hence, ∠BAC = 50°.

We know that,

In a cyclic quadrilateral, the sum of opposite angles is 180°.

⇒ ∠QPS + ∠QRS = 180°

⇒ 73° + ∠QRS = 180°

⇒ ∠QRS = 180° - 73°

⇒ ∠QRS = 107°.

From figure,

⇒ ∠PSR + ∠PQR = 180°

⇒ ∠PSR + ∠PQS + ∠RQS = 180°

⇒ 82° + 55° + ∠RQS = 180°

⇒ 137° + ∠RQS = 180°

⇒ ∠RQS = 180° - 137°

⇒ ∠RQS = 43°.

By angle sum property of a triangle we get,

⇒ ∠PSQ + ∠PQS + ∠QPS = 180°

⇒ ∠PSQ + 55° + 73° = 180°

⇒ ∠PSQ + 128° = 180°

⇒ ∠PSQ = 180° - 128°

⇒ ∠PSQ = 52°.

∠PSQ = ∠PRQ = 52° [Angles in the same segment]

Hence, ∠QRS = 107°, ∠RQS = 43° and ∠PRQ = 52°.

(i) Since ABC is an equilateral triangle so,

∠A = ∠B = ∠C = 60°.

From figure,

∠BDC = ∠BAC (Angles in same segment are equal.)

∴ ∠BDC = 60°.

Hence, ∠BDC = 60°.

(ii) BDCE is a cyclic quadrilateral. Hence, sum of the opposite angles = 180°.

⇒ ∠BDC + ∠BEC = 180°

⇒ 60° + ∠BEC = 180°

⇒ ∠BEC = 180° - 60° = 120°.

Hence, ∠BEC = 120°.

We know that,

Angle at the center is double the angle at the circumference subtended by the same chord.

∠AOC = 2∠ABC

∠AOC = 2x

∠x = $\frac{1}{2}$ ∠AOC = $\frac{160^{\circ}}{2}$ = 80°.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

In quadrilateral ABCD,

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠x + ∠y = 180°

⇒ 80° + ∠y = 180°

⇒ ∠y = 100°.

Substitute values in L.H.S of equation 3∠y − 2∠x = 140° :

= 3(100°) − 2(80°)

= 300° − 160°

= 140°.

As, L.H.S = R.H.S

Hence, proved that 3∠y − 2∠x = 140°.

From figure,

⇒ ∠CPB + ∠APB = 180° [Linear pair]

⇒ ∠CPB = 180° - ∠APB

⇒ ∠CPB = (180° − 120°) = 60°.

By angle sum property of a triangle we get,

⇒ ∠PCB + ∠CPB + ∠PBC = 180°

⇒ ∠PCB + 60° + 25° = 180°

⇒ ∠PCB + 85° = 180°

⇒ ∠PCB = 180° - 85°

⇒ ∠PCB = 95°

⇒ ∠ADB = ∠ACB = 95° [Angles in same segment of a circle are equal]

Hence, the value of ∠ADB = 95°.

(i) As equal chords of a circle subtend equal angles at the center and chord BC = chord CD, so ∠BOC = ∠COD.

∠BOD = 2 × ∠BAD

∠BOD = 2 × 75°

∠BOD = 150°

∠BOC = $\frac{1}{2}$ ∠BOD

= $\frac{1}{2}$ × 150° = 75°.

Hence, the value of ∠BOC = 75°.

(ii) Join BD.

Since, OB = OD

∴ ∠OBD = ∠ODB = x

Since sum of angles of triangle = 180°

In △OBD

⇒ ∠BOD + ∠OBD + ∠ODB = 180°

⇒ 150° + x + x = 180°

⇒ 150° + 2x = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = 15°.

Hence, the value of ∠OBD = 15°.

(iii) ABCD is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 75° = 180°

⇒ ∠BCD = 180° - 75°

⇒ ∠BCD = 105°.

Hence, the value of ∠BCD = 105°.

(i) Given,

∠DAE = 70°

From figure,

⇒ ∠DAE + ∠BAD = 180° [Linear pairs]

⇒ 70° + ∠BAD = 180°

⇒ ∠BAD = 180° - 70° = 110°.

We know that,

Sum of opposite angles in a cyclic quadrilateral = 180°

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 110° = 180°

⇒ ∠BCD = 180° - 110° = 70°.

Hence, ∠BCD = 70°.

(ii) We know that,

Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠BOD = 2∠BCD = 2 × 70° = 140°.

Hence, ∠BOD = 140°.

(iii) In △OBD,

OB = OD [Radius of same circle]

∠OBD = ∠ODB = x.

⇒ ∠OBD + ∠ODB + ∠BOD = 180°

⇒ x + x + 140° = 180°

⇒ 2x = 180° - 140°

⇒ 2x = 40°

⇒ x = $\frac{40^{\circ}}{2}$ = 20°.

∴ ∠OBD = 20°.

Hence, ∠OBD = 20°.

From figure,

∠AOC + ∠BOC = 180° [Linear pair]

100° + ∠BOC = 180°

∠BOC = 180° - 100°

∠BOC = 80°.

We know that,

Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠BOC = 2∠BDC

⇒ 80° = 2∠BDC

⇒ ∠BDC = $\frac{80^{\circ}}{2}$ = 40°

Hence, ∠BDC = 40°.

From the figure,

∠ABC + ∠CBE = 180° [Linear pair]

∠ABC + 110° = 180°

∠ABC = 70°

For the points A, B, C, D to be concyclic, ∠ABC + ∠ADC must be equal to 180°.

(i) When x = 70

x° + ∠ADC = 180° [Linear pair]

∠ADC = 180° − x° = 180° - 70° = 110°.

∠ABC + ∠ADC = 70° + 110° = 180°

Since the sum of opposite angles is 180°, the points A, B, C, D are concyclic when x = 70.

Hence, yes the points A, B, C, D are concyclic when x = 70.

(ii) When x = 80

x° + ∠ADC = 180° [Linear pair]

∠ADC = 180° − 80° = 100°

Check for con-cyclicity,

∠ABC + ∠ADC = 70° + 100° = 170°

Since the sum of opposite angles is not 180°, the points A, B, C, D are not concyclic when x = 80.

Hence, no the points A, B, C, D are not concyclic when x = 80.

(i) We know that,

Sum of opposite angles in a cyclic quadrilateral = 180°.

In cyclic quadrilateral ABCD,

∴ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 65° = 180°

⇒ ∠BCD = 180° - 65° = 115°.

Hence, ∠BCD = 115°.

(ii) In △ABD,

⇒ ∠ADB + ∠BAD + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠ADB + 65° + 70° = 180°

⇒ ∠ADB + 135° = 180°

⇒ ∠ADB = 180° - 135° = 45°.

We know that,

Angles in same segment are equal.

∴ ∠ACB = ∠ADB = 45°.

From figure,

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Since, angle in a semi-circle is a right angle. Thus, AC is the diameter.

Hence, ∠ADB = 45° and proved that AC is a diameter.

(i) We know that,

Angles in same segment are equal.

∠CBD = ∠CAD = 25°

Hence, ∠CBD = 25°.

(ii) We know that,

Angles in same segment are equal.Therefore,

⇒ ∠ACD = ∠ABD = 50°

⇒ ∠ACB = ∠ADB = 35°

From figure

⇒ ∠BCD = ∠ACD + ∠ACB

⇒ ∠BCD = 35° + 50°

⇒ ∠BCD = 85°.

We know that,

Sum of opposite angles in a cyclic quadrilateral = 180°.

⇒ ∠DAB + ∠DCB = 180°

⇒ ∠DAB = 180° - 85°

⇒ ∠DAB = 95°.

From figure,

⇒ ∠CAB = ∠DAB - ∠DAC

= 95° - 25°

= 70°.

Hence, ∠CAB = 70°.

(iii) Angles in same segment are equal. Therefore,

∠ACB = ∠ADB = 35°

Hence, ∠ACB = 35°.

(i) We know that,

Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.

∠BAD = Exterior ∠BCE = 80°.

From figure,

∠CAD = ∠BAD - ∠BAC = 80° - 25° = 55°.

Hence, ∠CAD = 55°.

(ii) We know that,

Angles in same segment are equal.

∴ ∠CBD = ∠CAD = 55°.

Hence, ∠CBD = 55°.

(iii) Since, AB ∥ DC,

∠ACD = ∠BAC = 25° [Alternate angles are equal]

In triangle ADC,

By angle sum property of triangle,

⇒ ∠ADC + ∠CAD + ∠ACD = 180°

⇒ ∠ADC + 55° + 25° = 180°

⇒ ∠ADC + 80° = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°.

Hence, ∠ADC = 100°.

(i) Given,

∠BED = 70°

We know that,

Angles in the same segment of a circle are equal.

∠BAD = ∠BED = 70°.

Hence, ∠BAD = 70°.

(ii) We know that,

The angle which an arc of a circle subtends at the center is double which it subtends at any point on the remaining part of the circumference.

Therefore,

∠BOD = 2∠BAD = 2 × 70° = 140°.

Hence, ∠BOD = 140°.

(iii) We know that,

Angle in a semi-circle is a right angle triangle.

∠ABD = 90°

In △ABD,

⇒ ∠BAD + ∠BDA + ∠ABD = 180°

⇒ 70° + ∠BDA + 90° = 180°

⇒ 160° + ∠BDA = 180°

⇒ ∠BDA = 180° - 160°

⇒ ∠BDA = 20°

From figure,

BC || AD

∴ ∠DBC = ∠BDA = 20° (Alternate angles are equal).

Hence, ∠DBC = 20°.

(iv) The figure ABCD is a cyclic quadrilateral

We know that,

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

∠DCF = ∠BAD = 70°.

Hence, ∠DCF = 70°.

(i) We know that,

Angle in semi-circle is a right angle.

∴ ∠AEB = 90°.

In △AEB,

⇒ ∠AEB + ∠EBA + ∠EAB = 180°

⇒ 90° + ∠EBA + 65° = 180°

⇒ 155° + ∠EBA = 180°

⇒ ∠EBA = 180° - 155° = 25°.

Hence, ∠EBA = 25°.

(ii) ∠BED = ∠EBA = 25° Alternate interior angles ED ∥ AB, EB as a transversal

Hence, ∠BED = 25°.

(iii) As, AB ∥ ED

∴ ∠DEB = ∠EBA = 25° [Alternate angles]

BCDE is a cyclic quadrilateral.

∴ ∠DEB + ∠BCD = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ 25° + ∠BCD = 180°

⇒ ∠BCD = 180° - 25° = 155°.

Hence, ∠BCD = 155°.

(i) We know that,

In a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.

∠ADC = ∠CBE = 55°

Hence, ∠ADC = 55°.

(ii) From figure,

∠CBE + ∠ABC = 180° [Linear pairs]

∠ABC = 180° - 55°

∠ABC = 125°.

Hence, ∠ABC = 125°.

(iii) We know that,

Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

∠AOC = 2∠ADC

∠AOC = 2(55°)

∠AOC = 110°

The reflex angle AOC,

x° = 360° - angle AOC

x° = 360° - 110°

x° = 250°.

Hence, x = 250.

We know that,

Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.

∠EDC = ∠A ....(i)

∠DCE = ∠B ....(ii)

AB ∥ CD

∠EDC = ∠B [Corresponding angles] ......(iii)

∠DCE = ∠A [Corresponding angles] .......(iv)

From (i), (ii), (iii) and (iv) we get :

∴ ∠A = ∠B

BE = AE [Sides opposite to equal angles are equal]

Hence, proved that ΔAEB is isosceles.

PQAD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠DAQ + ∠DPQ = 180°

⇒ 80° + ∠DPQ = 180°

⇒ ∠DPQ = 180° - 80°

⇒ ∠DPQ = 100°.

Also,

⇒ ∠PDA + ∠PQA = 180°

⇒ 84° + ∠PQA = 180°

⇒ ∠PQA = 180° - 84°

⇒ ∠PQA = 96°.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

(i) ∠QBC = ∠DPQ = 100°.

Hence, ∠QBC = 100°.

(ii) ∠BCP = ∠PQA = 96°.

Hence, ∠BCP = 96°.

(i) We know that,

An Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. ∠EDB is the exterior angle at vertex D. The interior opposite angle is ∠CAB.

⇒ ∠EBD = ∠CAB = 50°

Hence, the value of ∠EDB = 50°.

(ii) We know that,

⇒ OD = OB [radii of same circle]

From figure,

⇒ ∠AOD + ∠DOB = 180° [Linear pairs]

⇒ 140° + ∠DOB = 180°

⇒ ∠DOB = 180° - 140°

⇒ ∠DOB = 40°

In ΔODB,

By angle sum property of triangle,

⇒ ∠DOB + ∠ODB + ∠OBD = 180°

⇒ 40° + 2∠OBD = 180°

⇒ 2∠OBD = 180° - 40°

⇒ 2∠OBD = 140°

⇒ ∠OBD = $\frac{180^{\circ}}{2}$

⇒ ∠OBD = 70°

From figure,

⇒ ∠OBD + ∠EBD = 180° [Linear pairs]

⇒ 70° + ∠EBD = 180°

⇒ ∠EBD = 180° - 70°

⇒ ∠EBD = 110°

Hence, the value of ∠EBD = 110°.

(i) We know that,

Angles in same segment are equal.

∠DCB = ∠BAD = 35°

Hence, the value of ∠DCB = 35°.

(ii) From figure,

∠ADB = 90° (∵ Angle in semicircle is 90°.)

∴ ∠BDF = 90°

In △DBF,

⇒ ∠DBF + ∠DFB + ∠BDF = 180° (∵ By angle sum triangle property).

⇒ ∠DBF = 180° - (∠DFB + ∠BDF)

⇒ ∠DBF = 180° - (90° + 25°)

⇒ ∠DBF = 65°.

⇒ ∠DBF + ∠DBC = 180° (Linear pairs)

⇒ ∠DBC = 180° - 65°

⇒ ∠DBC = 115°.

Hence, the value of ∠DBC = 115°.

(iii) In △DCB,

⇒ ∠DCB + ∠CBD + ∠BDC = 180° (∵ By angle sum triangle property).

⇒ ∠BDC = 180° - (∠DCB + ∠CBD)

⇒ ∠BDC = 180° - (35° + 115°)

⇒ ∠BDC = 180° - (35° + 115°)

⇒ ∠BDC = 30°.

Hence, the value of ∠BDC = 30°.

(i) Since, AB and CD pass through the center, thus they are the diameters of circle.

In △CED,

∠CED = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠CED + ∠DCE + ∠CDE = 180°.

⇒ 90° + 40° + ∠CDE = 180°

⇒ ∠CDE + 130° = 180°

⇒ ∠CDE = 180° - 130°

⇒ ∠CDE = 50°.

Hence, ∠CDE = 50°.

(ii) From figure,

⇒ ∠AOD + ∠DOB = 180° (Linear pairs)

⇒ 75° + ∠DOB = 180°

⇒ ∠DOB = 180° - 75°

⇒ ∠DOB = 105°.

In △DOB,

∠ODB = ∠CDE = 50°

We know that,

Sum of angles of a triangle is 180°.

⇒ ∠DOB + ∠ODB + ∠DBO = 180°.

⇒ 105° + 50° + ∠DBO = 180°

⇒ ∠DBO + 155° = 180°

⇒ ∠DBO = 180° - 155°

⇒ ∠DBO = 25°.

From figure,

∠OBE = ∠DBO

∴ ∠OBE = 25°.

Hence, ∠OBE = 25°.

(i) Given,

AC = CD

In triangle ACD,

∠DAC = ∠ADC = 35° [Angles opposite to equal sides in a triangle are equal]

In △ACD,

⇒ ∠DAC + ∠ADC + ∠ACD = 180°

⇒ 35° + 35° + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 180° - 70° = 110°.

From figure,

⇒ ∠ACB + ∠ACD = 180° [Linear pair]

⇒ ∠ACB + 110° = 180°

⇒ ∠ACB = 180° - 110° = 70°.

Given,

AB = AC

∴ ∠ABC = ∠ACB = 70°. [As angles opposite to equal sides are equal]

Hence, ∠ABC = 70°.

(ii) In △ABC,

⇒ ∠BAC + ∠ACB + ∠ABC = 180° [Angle sum property of triangle]

⇒ ∠BAC + 70° + 70° = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° - 140° = 40°.

We know that,

Angles in same segment are equal.

⇒ ∠BEC = ∠BAC = 40°.

Hence, ∠BEC = 40°.

We know that,

An exterior angle of a triangle equals to the sum of the two interior opposite angles.

∠CBD = ∠A + ∠C

∠BCE = ∠A + ∠B.

Since BP bisects the exterior angle ∠CBD :

∠PBC = $\frac{1}{2}$ ∠CBD

Similarly, since CP bisects the exterior angle ∠BCE :

∠PCB = $\frac{1}{2}$ ∠BCE

In △BPC,

By angle sum property of triangle,

$$\begin{aligned}\Rightarrow ∠PBC + ∠PCB + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \frac{1}{2}∠CBD + \frac{1}{2}∠BCE + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \frac{1}{2}(∠A + ∠C) + \frac{1}{2}(∠A + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow ∠A + \frac{1}{2}(∠C + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \frac{1}{2}(∠B + ∠C) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow \frac{1}{2}(180^{\circ} - A) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow 90^{\circ} - \frac{A}{2} + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - \Big[90^{\circ} - \frac{A}{2}\Big] \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - 90^{\circ} + \frac{A}{2}\\[1em] \Rightarrow ∠BPC = 90^{\circ} - \frac{A}{2}.\end{aligned}$$

For A, B, P, C to be concyclic we would need ∠A + ∠BPC = 180° But:

∵ ∠A + ∠BPC = ∠A + $90^{\circ} - \frac{∠A}{2} = 90^{\circ} + \frac{∠A}{2}$

Hence, proved that $∠BPC = 90° - \frac{∠A}{2}$ and ABPC is not cyclic.

Given,

⇒ QR = OP

⇒ QR = OQ (As, OP = OQ = Radius of circle)

⇒ ∠QOR = ∠ORQ = 20° (Angles opposite to equal sides of a triangle are equal)

Exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ ∠OQP = ∠QOR + ∠ORQ = 20° + 20° = 40°.

As OP = OQ, ∠OPQ = ∠OQP

⇒ ∠OPQ = 40°

⇒ ∠OPR = 40°.

Exterior angle in a triangle is equal to the sum of two opposite interior angles.

In triangle OPR,

∴ x° = ∠OPR + ∠ORP = 40° + 20° = 60°.

Hence, the value of x = 60.

(i) Join BI and CI.

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° (∵ sum of angles = 180°.)

⇒ ∠BAC + 55° + 65° = 180°

⇒ ∠BAC + 120° = 180°

⇒ ∠BAC = 180° - 120°

⇒ ∠BAC = 60°.

I is the incentre,

∴ I lies on the bisectors of angle of the △ABC,

∴ ∠BAD = ∠CAD = $\frac{60^{\circ}}{2}$ = 30°.

∠BCD = ∠BAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠BCD = 30°.

(ii) Similarly,

∠CBD = ∠CAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠CBD = 30°.

(iii) The line CI bisects ∠C (∵ I lies on the bisectors of angle of the △ABC).

∴ ∠BCI = $\frac{65^{\circ}}{2} = 32 \frac{1}{2}^{\circ}$.

From figure,

∠DCI = ∠BCD + ∠BCI = 30° + $32 \frac{1}{2}^{\circ}$

= 62.5°.

Hence, the value of ∠DCI = 62.5°.

(iv) ∠IBC = $\frac{55}{2}^{\circ}$ = 27.5°

∠ICB = $\frac{65}{2}^{\circ}$ = 32.5°

In triangle BIC,

By angle sum property of triangle,

∠BIC + ∠IBC + ∠ICB = 180°

∠BIC = 180° - (∠IBC + ∠ICB)

= 180° - (27.5° + 32.5°)

= 180° - 60°

= 120°.

Hence, the value of ∠BIC = 120°.

(i) Given that BD is a diameter of the circle.

We know that,

Angle in a semicircle is a right angle.

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

⇒ ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]

⇒ 58° + 90° + ∠BDC = 180°

⇒ 148° + ∠BDC = 180°

⇒ ∠BDC = 180° - 148° = 32°.

Hence, ∠BDC = 32°.

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral BECD

⇒ ∠BEC + ∠BDC = 180°

⇒ ∠BEC + 32° = 180°

⇒ ∠BEC = 180° - 32° = 148°

Hence, ∠BEC = 148°.

(iii) In cyclic quadrilateral ABEC,

⇒ ∠BAC + ∠BEC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠BAC + 148° = 180°

⇒ ∠BAC = 180° - 148° = 32°.

Hence, ∠BAC = 32°.

(i) ∠ABC = 90° [Angle in semicircle]

In ∆ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180° [Angle sum property of triangle]

⇒ 90° + 50° + ∠ACB = 180°

⇒ 140° + ∠ACB = 180°

⇒ ∠ACB = 40°

Hence, ∠ACB = 40°.

(ii) ∠EAC = ∠ACB = 40° [Alternate angles, AC transversal to parallel lines AE and BC]

∠EAC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

40° + ∠EDC = 180°

∠EDC = 140°

Hence, ∠EDC = 140°.

(iii) ∠EBC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

140° + ∠EBC = 180°

∠EBC = 40°

In ∆EBC,

⇒ ∠BEC + ∠ECB + ∠EBC = 180° [Angle sum property of triangle]

⇒ ∠BEC + 90° + 40° = 180°

⇒ ∠BEC = 180° - 130°

⇒ ∠BEC = 50°

In ∆EAB,

⇒ ∠EAB = ∠EAC + ∠BAC

= 40° + 50° = 90°

We know that, if an angle of a triangle in a circle is 90° then, the hypotenuse must be the diameter of the circle.

Hence, ∠BEC = 50° and BE is the diameter of the circle.

(i) We know that,

Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

∠ABC = $\frac{1}{2}$ ∠AOC

∠ABC = $\frac{72^{\circ}}{2}$ = 36°

Hence, ∠ABC = 36°.

(ii) Given,

AC = BD

∠BAC = ∠ABC = 36° [Angles opposite to equal sides are equal]

Hence, ∠BAC = 36°.

(iii) In ∆ABD,

∠ABD + ∠BAD + ∠ADB = 180° [Angle sum property of triangle]

∠ABD + 36° + 90° = 180°

∠ABD + 126° = 180°

∠ABD = 180° - 126°

∠ABD = 54°

Hence, ∠ABD = 54°.

(i) From figure,

⇒ ∠ADC = ∠ABC (Angles in same segment are equal)

⇒ ∠ADC = 62°.

Hence, ∠ADC = 62°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ACB = 90°

Using angle sum property,

⇒ ∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 90° + 62° = 180°

⇒ ∠CAB + 152° = 180°

⇒ ∠CAB = 180° - 152°

⇒ ∠CAB = 28°.

Hence, ∠CAB = 28°.

An angle subtended by an arc at a point on the remaining part of the circle is an angle at the circumference.

An angle in a semicircle is a right angle.

Hence, option 3 is the correct option.

The sum of the opposite angles of a cyclic quadrilateral is 180°.

Hence, option 3 is the correct option.

If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is a cyclic quadrilateral.

Hence, option 4 is the correct option.

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Hence, option 2 is the correct option.

Any two angles formed in the same segment of a circle are equal.

Hence, option 4 is the correct option.

Given,

ABCD is a cyclic parallelogram.

∠ABC = ∠ADC [Opposite angles of parallelogram are equal]

∠ABC + ∠ADC = 180° [Sum of opposite angles of cyclic quadrilateral is 180°]

∠ABC + ∠ABC = 180°

2∠ABC = 180°

∠ABC = $\frac{180^{\circ}}{2}$

∠ABC = ∠ADC = 90°

A parallelogram one of whose angle is 90° is a rectangle.

Hence, Every cyclic parallelogram is a rectangle.

Hence, option 3 is the correct option.

Given,

ABCD is an isosceles trapezium with AB ∥ CD and AD = BC.

∠A + ∠D = 180° [Interior angles on the same side of a transversal are supplementary]

In an isosceles trapezium,

∠A = ∠B and ∠C = ∠D [Base angles are equal]

So,

∠A + ∠C = 180°

Since a pair of opposite angles of quadrilateral ABCD is supplementary,

ABCD is a cyclic quadrilateral.

Hence, an isosceles trapezium is always a cyclic quadrilateral.

Hence, option 4 is the correct option.

Given,

ABCD is a rhombus.

In a rhombus,

All sides are equal, but angles are not necessarily equal to 90°.

Opposite angles of a rhombus are equal.

So,

∠A = ∠C and ∠B = ∠D

But,

∠A + ∠C ≠ 180° (always)

Hence, opposite angles of a rhombus are not always supplementary.

Therefore, a rhombus is not always a cyclic quadrilateral.

Hence, option 2 is the correct option.

Given,

ABCD is a cyclic quadrilateral in which AP, BP, CR and DR are the angle bisectors of ∠A, ∠B, ∠C and ∠D respectively, forming quadrilateral PQRS.

In ΔPAB,

∠APB + ∠PAB + ∠PBA = 180° [Sum of the angles of a triangle is 180°]

But,

∠PAB = $\frac{1}{2}$∠A and ∠PBA = $\frac{1}{2}$∠B [AP and BP are angle bisectors]

So,

∠APB + $\frac{1}{2}$∠A + $\frac{1}{2}$∠B = 180° …(i)

Similarly, in ΔRCD,

∠CRD + ∠RCD + ∠RDC = 180° [Sum of the angles of a triangle is 180°]

But,

∠RCD = $\frac{1}{2}$∠C and ∠RDC = $\frac{1}{2}$∠D [CR and DR are angle bisectors]

So,

∠CRD + $\frac{1}{2}$∠C + $\frac{1}{2}$∠D = 180° …(ii)

Adding (i) and (ii),

∠APB + ∠CRD + $\frac{1}{2}$ (∠A + ∠B + ∠C + ∠D) = 360°

But,

∠A + ∠B + ∠C + ∠D = 360° [Sum of angles of a quadrilateral]

So,

∠APB + ∠CRD + $\frac{1}{2}$ × 360° = 360°

∠APB + ∠CRD + 180° = 360°

∠APB + ∠CRD = 180°

Since a pair of opposite angles of quadrilateral PQRS is supplementary,

PQRS is a cyclic quadrilateral.

Hence, option 1 is the correct option.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠AOB = 2∠ACB

⇒ ∠AOB = 2 × 30°

⇒ ∠AOB = 60°.

Hence, option 3 is the correct option.

Given,

Since O is the centre of the circle, OA and OB are both radii. OA = OB Therefore,

∠OBA = ∠OAB = 35° [Angles opposite to equal sides of a triangle are equal]

In ΔAOB,

By angle sum property of triangle,

∠OBA + ∠OAB + ∠AOB = 180°

35° + 35° + ∠AOB = 180°

∠AOB = 180° - 35° - 35°

∠AOB = 110°.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠AOB = 2∠ACB

⇒ 110° = 2∠ACB

⇒ ∠ACB = $\frac{110^{\circ}}{2}$

⇒ ∠ACB = 55°.

Hence, option 2 is the correct option.

Given,

Points A and D are on the circle, and both form angles with the chord BC.

∠BDC = ∠BAC = 60° [Angles in same segment are equal]

In ΔBCD,

By angle sum property of triangle,

∠BCD + ∠DBC + ∠BDC = 180°

∠BCD = 180° - ∠DBC - ∠BDC

∠BCD = 180° - 40° - 60°

∠BCD = 80°.

Hence, option 4 is the correct option.

In ΔOAC,

OA = OC [Radii of same circle]

Since, two sides are equal it is isosceles triangle and opposite sides are also equal:

∠OCA = ∠OAC = 55°

In ΔOBD,

OB = OD [Radii of same circle]

Since, two sides are equal it is isosceles triangle and opposite sides are also equal:

∠OBD = ∠ODB

∠AOC = ∠BOD [vertically opposite angles]

In ΔOAC,

By angle sum property of triangle,

∠AOC + ∠OAC + ∠OCA = 180°

∠AOC + 55° + 55° = 180°

∠AOC + 110° = 180°

∠AOC = 180° - 110°

∠AOC = 70°

∠AOC = ∠BOD = 70°

In ΔOBD,

By angle sum property of triangle,

∠BOD + ∠OBD + ∠ODB = 180°

70° + 2∠OBD = 180°

2∠OBD = 180° - 70°

2∠OBD = 110°

∠OBD = $\frac{110^{\circ}}{2}$

∠OBD = 55°.

Hence, option 1 is the correct option.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠OQA = 90°

In △QAP,

⇒ ∠PQA + ∠QAP + ∠APQ = 180°

⇒ 90° + ∠QAP + 20° = 180°

⇒ ∠QAP = 180° - 90° - 20° = 70°.

In △OPA,

⇒ OA = OP (Radii of same circle)

⇒ ∠OAP = ∠OPA (Angle opposite to equal sides are equal)

⇒ ∠OAP = 20°.

From figure,

⇒ ∠OAQ = ∠QAP - ∠OAP = 70° - 20° = 50°.

Hence, Option 3 is the correct option.

Join BC.

In △BOC,

Since,

OB = OC (Radius of same circle)

∴ ∠OBC = ∠OCB = x (let)

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + ∠BOC = 180°

⇒ ∠BOC = 180° - 2x

In △AOC,

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ ∠BAC + (30° + x) + (40° + x) = 180°

⇒ ∠BAC + 70° + 2x = 180°

⇒ ∠BAC = 180° - 70° - 2x

⇒ ∠BAC = 110° - 2x

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠BOC = 2∠BAC

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ 4x - 2x = 220° - 180°

⇒ 2x = 40°

⇒ x = $\frac{40^{\circ}}{2}$ = 20°

⇒ ∠BOC = 180° - 2x

⇒ ∠BOC = 180° - 2(20°) = 180° - 40° = 140°.

Hence, option 4 is the correct option.

From figure,

∠BOC = 360° - (∠AOB + ∠AOC)

∠BOC = 360° - (110° + 80°)

= 360° - 190°

= 170°.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ BOC = 2∠BAC

∴ ∠BAC = $\frac{1}{2}$ ∠BOC

= $\frac{170^{\circ}}{2}$ = 85°.

Hence, option 3 is the correct option.

From figure,

OP = OQ (Radii of same circle)

∠OQP = ∠OPQ = 30° (As angles opposite to equal sides are equal)

In △POQ,

By angle sum property of triangle,

∠POQ + ∠OQP + ∠OPQ = 180°

∠POQ = 180° - ∠OQP - ∠OPQ

∠POQ = 180° - 30° - 30°

∠POQ = 120°.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠POQ = 2∠PAQ

∴ ∠PAQ = $\frac{1}{2}$ ∠POQ

= $\frac{120^{\circ}}{2}$ = 60°.

Hence, option 1 is the correct option.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠ACB = $\frac{1}{2}$ Reflex∠AOB

= $\frac{360^{\circ} - 150^{\circ}}{2}$ = 105°.

Hence, option 2 is the correct option.

The sum of the opposite angles of a cyclic quadrilateral is 180°.

∠A + ∠C = 180°

y + 5y = 180°

6y = 180°

y = 30°

∠B + ∠D = 180°

3x + x = 180°

4x = 180°

x = 45°

∠A = y = 30°

∠C = 5y = 150°

∠D = x = 45°

∠B = 3x = 135°

Hence, option 3 is the correct option.

From figure,

∠ACD and ∠ACB forms linear pairs [BD is a straight line].

⇒ ∠ACD + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180° - 120°

⇒ ∠ACB = 60°.

In ΔABC,

⇒ BC = AC [Given]

⇒ ∠ABC = ∠BAC = x° (let) [Angles opposite to equal sides in triangle are equal]

According to angle sum property in ΔABC,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 60° + x° + x° = 180°

⇒ 2x° = 180° - 60°

⇒ 2x° = 120°

⇒ x° = $\frac{120°}{2}$ = 60°.

From figure,

ABCE is cyclic quadrilateral.

We know that,

Opposite angles of cyclic quadrilateral are supplementary.

⇒ ∠ABC + ∠AEC = 180°

⇒ 60° + ∠AEC = 180°

⇒ ∠AEC = 180° - 60°

⇒ ∠AEC = 120°.

Hence, option 4 is the correct option.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠POR = 2∠PQR = 2 × 35° = 70°.

Hence, option 4 is the correct option.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠BOC = 2∠BDC

∠BOC = 2(36°) = 72°

∠AOB + ∠BOC = 180°

∠AOB = 180° - ∠BOC

∠AOB = 180°- 72° = 108°

∴ ∠ACB = $\frac{1}{2}$ ∠AOB

∠ACB = $\frac{108^{\circ}}{2}$ = 54°.

Hence, option 2 is the correct option.

∠ADC = 90° and ∠ADC = 90° [Angle in semicircle is a right angle]

In △ABC,

∠BAC = 58°

∠ABC = 90°

By angle sum property of triangle,

∠BCA + ∠BAC + ∠ABC = 180°

∠BCA = 180° - (∠BAC + ∠ABC)

∠BCA = 180° - (90° + 58°) = 32°

In △ABC :

∠DAC = 65° and ∠ADC = 90°

By angle sum property of triangle,

∠DCA + ∠DAC + ∠ADC = 180°

∠DCA = 180° - (∠DAC + ∠ADC)

∠DCA = 180° - (90° + 65°) = 25°

From figure,

∠BCD = ∠DCA + ∠BCA = 32° + 25° = 57°.

Hence, option 1 is the correct option.

A minor segment is the smaller of the two regions created when a chord divides a circle. The corresponding arc is the minor arc.

The angle formed in a minor segment of a circle is an obtuse angle.

Hence, option 2 is the correct option.

A major segment is the larger of the two regions created when a chord divides a circle. The corresponding arc is the major arc.

The angle formed in a major segment of a circle is an acute angle.

Hence, option 1 is the correct option.

In equilateral triangle ABC,

∠BAC = 60°

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠BOC = 2∠BAC = 2(60°) = 120°.

Hence, option 1 is the correct option.

Let the two diameters AB and CD intersect at the centre O at right angles.

The quadrilateral formed by joining the end points in order is ACBD.

In ΔAOC, ΔCOB, ΔBOD, ΔDOA , we have two sides which are radii and an included angles as 90°. Therefore, by SAS theorem, all triangles are congruent.

Thus, all four sides are equal AC = CB = BD = DA.

Each interior angle of the quadrilateral (e.g., ∠ACB) is an angle inscribed in a semicircle. Thus, each interior angle equals to 90°.

A quadrilateral with all sides equal, and all angles equal to 90° is a square.

Hence, option 2 is the correct option.

Reflex ∠AOC = 360° - 130° = 230°

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠ABC = $\frac{1}{2}$ Reflex ∠AOC

∠ABC = $\frac{230^{\circ}}{2}$ = 115°.

Hence, option 4 is the correct option.

If chord length AB = radius(OA, OB), then the triangle OAB formed by two radii and the chord is equilateral.

Central angle subtended by the chord = ∠AOB = 60°

∠ACB = $\frac{1}{2}$ ∠AOB = $\frac{1}{2}$ (60°) = 30°.

Opposite angles of a cyclic quadrilateral ADBC are supplementary.

∠ACB + ∠ADB = 180°

∠ADB = 180° - 30°

∠ADB = 150°.

Thus, angle subtended by this chord at the minor arc of the circle is 150°.

Hence, option 4 is the correct option.

∠ACB = ∠ADB = 45° [Angles in the same segment]

From figure,

∠BCD = ∠ACD + ∠ACB

∠BCD = 55° + 45° = 100°

In a cyclic quadrilateral, the sum of opposite angles is 180°.

∠DAB + ∠BCD = 180°

∠DAB = 180° - ∠BCD

∠DAB = 180° - 100° = 80°.

Hence, option 4 is the correct option.

∠ABD = ∠ACD = 30° [Angles in the same segment]

From figure,

∠PBD = ∠ABD = 30°

In ΔPBD,

By angle sum property,

∠PBD + ∠BPD + ∠PDB = 180°

30° + 100° + ∠PDB = 180°

∠PDB = 180° - 130°

∠PDB = 50°.

From figure,

∠CDB = ∠PDB = 50°.

Hence, option 3 is the correct option.

Opposite angles of a cyclic quadrilateral are supplementary.

So,

∠ABC + ∠ADC = 180°

∠ABC = 180° − 125° = 55°.

∠ACB = 90° [Angle in semicircle is a right angle]

In △ABC,

By angle sum property of triangle,

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + 55° + 90° = 180°

∠BAC = 180° − 145° = 35°.

Hence, option 1 is the correct option.

From figure,

∠DAB = 90° [Angle in the semicircle is a right angle]

In the figure, both ∠ADB and ∠ACB are subtended by the same arc AB. Therefore:

∠ADB = ∠ACB = 2x

In triangle ADB,

∠DAB + ∠ADB + ∠ABD = 180°

90° + 2x + 3x = 180°

5x = 90°

x = 18°.

Hence, option 2 is the correct option.

If the length of chord AB is equal to the radius of the circle; then △AOB is equilateral triangle.

∠AOB = 60°

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠ACB = $\frac{1}{2}$ ∠AOB

∠ACB = $\frac{60^{\circ}}{2}$ = 30°.

Hence, option 1 is the correct option.

Arc AD subtends ∠ABD and ∠ACD. Therefore,

∠ABD = ∠ACD = 55° [From Figure]

Arc AB subtends ∠BCA and ∠BDA. Therefore,

∠BCA = ∠BDA = 32° [From Figure]

From figure,

∠BCD = ∠BCA + ∠ACD

∠BCD = 32° + 55°

∠BCD = 87°.

Hence, option 3 is the correct option.

Join AC and BD.

∠BDC = 90° (Angle in semicircle is right angle)

OA = OB (Radii of same circle)

∠OBA = ∠OAB = 50° (Angles opposite to equal sides are equal in a triangle)

In triangle OAB,

∠AOB + ∠OAB + ∠OBA = 180°

∠AOB = 180° - (∠OAB + ∠OBA)

∠AOB = 180° - (50° + 50°)

∠AOB = 80°.

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∠ACB = $\frac{1}{2}$ ∠AOB

∠ACB = $\frac{80°}{2}$

∠ACB = 40°

∠ADB = ∠ACB = 40° [Angle in same segment are equal]

From figure,

∠BDC = ∠ADB + x

90° = 40° + x

x = 50°.

Hence, option 2 is the correct option.

Join AD.

∠AOC = 130°

∠ADB = 90°

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠ADC = $\frac{1}{2}$ ∠AOC

∠ADC = $\frac{130^{\circ}}{2}$ = 65°

At point D,

∠ADB = ∠ADC + ∠CDB

90° = 65° + x

x = 25°

Hence, option 1 is the correct option.

We know that,

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠ACB = $\frac{1}{2}$ ∠AOB

∠ACB = $\frac{40^{\circ}}{2}$ = 20°

In ΔBDC, the sum of all angles is 180°

∠OBC + ∠BDC + ∠ACB = 180°

∠OBC + 100° + 20° = 180°

∠OBC = 180° - 120°

∠OBC = 60°

Hence, option 3 is the correct option.

In a cyclic quadrilateral, opposite angles are supplementary.

Since ∠BAD and ∠BCD are opposite angles,

(2x + 5)° + (x + 10)° = 180°

3x + 15 = 180°

3x = 165°

x = 55°

Hence, option 3 is the correct option.

From figure,

OA = OB (Radii of same circle)

∠OBA = ∠OAB = 50° (As angles opposite to equal sides in a triangle are equal)

∠OBA + ∠OAB + ∠AOB = 180° [Angle sum property of a triangle]

50° + 50° + ∠AOB = 180°

100° + ∠AOB = 180°

∠AOB = 180° - 100°

∠AOB = 80°.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

So, reason (R) is true.

∠AOB = 2∠ACB

∠ACB = $\frac{80°}{2}$ = 40°.

So, Assertion (A) is true.

Both A and R are true, and R correctly explains A.

Hence, option 1 is the correct option.

In triangle ABD,

∠DAB = 60°

∠DBA = 50°

From figure,

⇒ ∠DAB + ∠DBA + ∠ADB = 180° [Angle sum property of a triangle]

⇒ 60° + 50° + ∠ADB = 180°

⇒ 110° + ∠ADB = 180°

⇒ ∠ADB = 180° - 110°

⇒ ∠ADB = 70°

Angle in same segment are equal.

∠ADB = ∠ACB = 70°

So, assertion (A) is true.

We know that,

In cyclic quadrilateral opposite angles are supplementary (sum = 180°), not equal.

So, reason (R) is false.

A is true, R is false.

Hence, option 3 is the correct option.

In △ABC,

∠ACB = 90° [Angle in a semicircle]

Since ABCD is a cyclic quadrilateral, the sum of its opposite angles is 180°.

⇒ ∠ADC + ∠ABC = 180°

⇒ 130° + ∠ABC = 180°

⇒ ∠ABC = 50°

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° [Angle sum property of a triangle]

⇒ ∠BAC = 180° - (50° + 90°)

⇒ ∠BAC = 40°.

So, Assertion (A) is true.

Angles in the same segment of a circle are equal, so Reason (R) is a true statement. However, ∠BAC = 40° is obtained using the angle in a semicircle together with the cyclic-quadrilateral property, and not by the equality of angles in the same segment. So R is not the correct explanation of A.

Both A and R are true, but R is not the correct explanation of A.

Hence, option 2 is the correct option.

From figure,

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Reflex AOB = 360° - 100° = 260°

∠ACB = $\frac{1}{2}$ Reflex ∠AOB

∠ACB = $\frac{260 ^{\circ}}{2}$

∠ACB = 130°

∠ACB + ∠BCD = 180° [Linear pair]

∠BCD = 180° - 130°

∠BCD = 50°

So assertion (A) is false.

We know that,

In, cyclic quadrilaterals if one side is extended, the exterior angle formed is equal to the interior opposite angle.

So, reason (R) is true.

A is false, R is true

Hence, option 4 is the correct option.

(a) In △ BAE,

⇒ ∠AEB = 90° (Angle in a semicircle is a right angle)

By angle sum property of triangle,

⇒ ∠AEB + ∠EBA + ∠BAE = 180°

⇒ 90° + 33° + ∠BAE = 180°

⇒ ∠BAE + 123° = 180°

⇒ ∠BAE = 180° - 123° = 57°.

Hence, ∠BAE = 57°.

(b) From figure,

⇒ ∠EAC = ∠EAB + ∠BAC

⇒ 82° = 57° + ∠BAC

⇒ ∠BAC = 82° - 57° = 25°.

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

⇒ ∠BOC = 2∠BAC = 2 × 25° = 50°.

Hence, ∠BOC = 50°.

(c) From figure,

∠BOC and ∠BOD form a linear pair.

∴ ∠BOC + ∠BOD = 180°

⇒ 50° + ∠BOD = 180°

⇒ ∠BOD = 180° - 50° = 130°.

In △ BOD,

⇒ ∠BOD + ∠ODB + ∠DBO = 180°

⇒ 130° + ∠ODB + 33° = 180°

⇒ ∠ODB + 163° = 180°

⇒ ∠ODB = 180° - 163° = 17°.

Hence, ∠ODB = 17°.