Angle & Cyclic Properties of a Circle — Multiple Choice Questions

If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a :

1. minor arc

2. major arc

3. semi-circle

4. none of these

Answer

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 :

1. 90°

2. 150°

3. 180°

4. 360°

Answer

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 :

1. a rectangle

2. a parallelogram

3. a square

4. a cyclic quadrilateral

Answer

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 :

1. 90°

2. the interior opposite angle

3. the interior adjacent angle

4. any of the interior angles

Answer

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 :

1. complementary

2. supplementary

3. each right angle

4. equal

Answer

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

Hence, option 4 is the correct option.

Every cyclic parallelogram is a/an :

1. square

2. rhombus

3. rectangle

4. isosceles trapezium

Answer

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 = $\dfrac{180^{\circ}}{2}$

∠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.

An isosceles trapezium is always :

1. a parallelogram

2. a square

3. a rectangle

4. a cyclic quadrilateral

Answer

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.

Which of the following quadrilaterals is not always a cyclic quadrilateral?

1. square

2. rhombus

3. rectangle

4. an isosceles trapezium

Answer

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.

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is :

1. cyclic

2. square

3. rectangle

4. parallelogram

Answer

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

So,

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

Similarly, in ΔRCD,

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

But,

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

So,

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

Adding (i) and (ii),

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

But,

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

So,

∠APB + ∠CRD + $\dfrac{1}{2}$ × 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.

In the given figure, O is the centre of the circle and ∠ACB = 30°. Then, ∠AOB = ?

1. 15°

2. 30°

3. 60°

4. 90°

Answer

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.

In the given figure, O is the centre of the circle. If ∠OAB = 35° and C is a point on the circle, then ∠ACB = ?

1. 35°

2. 55°

3. 45°

4. 75°

Answer

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 = $\dfrac{110^{\circ}}{2}$

⇒ ∠ACB = 55°.

Hence, option 2 is the correct option.

In the given figure, ∠ABC and ∠DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 40°. Then, ∠BCD = ?

1. 60°

2. 40°

3. 100°

4. 80°

Answer

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 the given figure, O is the centre of the circle. If ∠OAC = 55°, then ∠OBD = ?

1. 55°

2. 35°

3. 45°

4. 70°

Answer

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 = $\dfrac{110^{\circ}}{2}$

∠OBD = 55°.

Hence, option 1 is the correct option.

In the adjoining figure, O is the center of the circle, and a semicircle is drawn on OA as the diameter. ∠APQ = 20°. The degree measure of ∠OAQ is :

1. 25°

2. 40°

3. 50°

4. 65°

Answer

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.

In the given figure, O is the centre of the circle in which ∠OBA = 30° and ∠OCA = 40°. Then, ∠BOC = ?

1. 70°

2. 100°

3. 120°

4. 140°

Answer

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 = $\dfrac{40^{\circ}}{2}$ = 20°

⇒ ∠BOC = 180° - 2x

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

Hence, option 4 is the correct option.

In the given figure, O is the centre of the circle.
If ∠AOB = 110° and ∠AOC = 80°, then ∠BAC = ?

1. 75°

2. 80°

3. 85°

4. 95°

Answer

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 = $\dfrac{1}{2}$ ∠BOC

= $\dfrac{170^{\circ}}{2}$ = 85°.

Hence, option 3 is the correct option.

In the given figure, O is the centre of the circle and ∠OPQ = 30°. Then, ∠PAQ = ?

1. 60°

2. 70°

3. 80°

4. 75°

Answer

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 = $\dfrac{1}{2}$ ∠POQ

= $\dfrac{120^{\circ}}{2}$ = 60°.

Hence, option 1 is the correct option.

In the given figure, O is the centre of the circle and ∠AOB = 150°. Then, ∠ACB = ?

1. 75°

2. 105°

3. 85°

4. 115°

Answer

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 = $\dfrac{1}{2}$ Reflex∠AOB

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

Hence, option 2 is the correct option.

In the given figure, ABCD is a cyclic quadrilateral. The measure of its greatest angle is :

1. 135°

2. 140°

3. 150°

4. 160°

Answer

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.

In the given diagram, chords AC and BC are equal. If ∠ACD = 120°, then ∠AEC is:

1. 30°

2. 60°

3. 90°

4. 120°

Answer

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° = $\dfrac{120°}{2}$ = 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.

In the given figure, O is the centre of the circle.
If ∠PQR = 35°, then ∠POR is equal to :

1. 17.5°

2. 35°

3. 60°

4. 70°

Answer

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.

In the given figure, O is the centre of the circle and ∠BDC = 36°. The measure of ∠ACB is :

1. 36°

2. 54°

3. 72°

4. 46°

Answer

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 = $\dfrac{1}{2}$ ∠AOB

∠ACB = $\dfrac{108^{\circ}}{2}$ = 54°.

Hence, option 2 is the correct option.

ABCD is a cyclic quadrilateral such that AC is a diameter of the circle. If ∠BAC = 58° and ∠DAC = 65°, then ∠BCD is equal to :

1. 57°

2. 123°

3. 90°

4. 60°

Answer

∠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.

The angle formed in a minor segment of a circle is :

1. an acute angle

2. an obtuse angle

3. a right angle

4. either an acute or an obtuse angle

Answer

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.

The angle formed in a major segment of a circle is :

1. an acute angle

2. an obtuse angle

3. a right angle

4. either an acute or an obtuse angle

Answer

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.

An equilateral triangle ABC is inscribed in a circle with centre O. The measure of ∠BOC is :

1. 120°

2. 60°

3. 90°

4. 30°

Answer

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.

If two diameters of a circle intersect each other at right angles, then the quadrilateral formed by joining their end points is a :

1. rhombus

2. square

3. rectangle

4. parallelogram

Answer

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.

In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC is equal to :

1. 65°

2. 90°

3. 105°

4. 115°

Answer

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 = $\dfrac{1}{2}$ Reflex ∠AOC

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

Hence, option 4 is the correct option.

A chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is :

1. 60°

2. 90°

3. 120°

4. 150°

Answer

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 = $\dfrac{1}{2}$ ∠AOB = $\dfrac{1}{2}$ (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.

ABCD is a cyclic quadrilateral such that ∠ADB = 45° and ∠DCA = 55°, then ∠DAB is equal to :

1. 55°

2. 45°

3. 100°

4. 80°

Answer

∠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.

In the given figure, if ∠ACD = 30° and ∠BPD = 100°, then ∠CDB is equal to :

1. 30°

2. 40°

3. 50°

4. 60°

Answer

∠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.

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 125°, then ∠BAC is equal to :

1. 35°

2. 40°

3. 45°

4. 55°

Answer

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.

In the given figure, if O is the centre of the circle, then the value of x is :

1. 15°

2. 18°

3. 20°

4. 24°

Answer

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.

In the given figure, O is the centre of the circle.
If the length of chord AB is equal to the radius of the circle; then ∠ACB is equal to :

1. 30°

2. 45°

3. 60°

4. 40°

Answer

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 = $\dfrac{1}{2}$ ∠AOB

∠ACB = $\dfrac{60^{\circ}}{2}$ = 30°.

Hence, option 1 is the correct option.

The value of x in the following figure is :

1. 80°

2. 57°

3. 87°

4. 96°

Answer

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.

In the given figure, if O is the centre of the circle, then the value of x is :

1. 40°

2. 50°

3. 70°

4. 30°

Answer

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 = $\dfrac{1}{2}$ ∠AOB

∠ACB = $\dfrac{80°}{2}$

∠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.

In the given figure, O is centre of the circle.
If ∠AOC = 130°, then the value of x is :

1. 25°

2. 50°

3. 65°

4. 40°

Answer

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 = $\dfrac{1}{2}$ ∠AOC

∠ADC = $\dfrac{130^{\circ}}{2}$ = 65°

At point D,

∠ADB = ∠ADC + ∠CDB

90° = 65° + x

x = 25°

Hence, option 1 is the correct option.

In the given figure, O is the centre of the circle, ∠AOB = 40° and ∠BDC = 100°. The measure of ∠OBC is :

1. 40°

2. 80°

3. 60°

4. 20°

Answer

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 = $\dfrac{1}{2}$ ∠AOB

∠ACB = $\dfrac{40^{\circ}}{2}$ = 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.

ABCD is a cyclic quadrilateral. If ∠BAD = (2x + 5)° and ∠BCD = (x + 10)°, then x is equal to :

1. 65°

2. 45°

3. 55°

4. 5°

Answer

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.