 # RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12B

## RS Aggarwal Solutions for Class 9 Maths Exercise 12B PDF

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## RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12B Download PDF                   ## Access RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12B

### Exercise 12(B) page: 456

1. (i) In Figure (1), O is the centre of the circle. If ∠OAB = 40o and ∠OCB = 30o, find ∠AOC.

(ii) In Figure (2), A, B and C are three points on the circle with centre O such that ∠AOB = 90o and

∠AOC = 110o. Find ∠BAC.  Solution:

(i) Join the line BO Consider △ BOC

We know that the sides are equal to the radius

So we get

OC = OB

From the figure we know that the base angles of an isosceles triangle are equal

∠OBC = ∠OCB

It is given that

∠OCB = 30o

So we get

∠OBC = ∠OCB = 30o

So we get ∠OBC = 30o ……. (1)

Consider △ BOA

We know that the sides are equal to the radius

So we get

OB = OC

From the figure we know that the base angles of an isosceles triangle are equal

∠OAB = ∠OBA

It is given that

∠OAB = 40o

So we get

∠OAB = ∠OBA = 40o

So we get ∠OBA = 40o ……. (2)

We know that

∠ABC = ∠OBC + ∠OBA

By substituting the values

∠ABC = 30o + 40o

So we get

∠ABC = 70o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

It can be written as

∠AOC = 2 × ∠ABC

So we get

∠AOC = 2 × 70o

By multiplication

∠AOC = 140o

Therefore, ∠AOC = 140o

(ii) From the figure we know that ∠AOB + ∠AOC + ∠BOC = 360o

By substituting the values

90o + 110o + ∠BOC = 360o

On further calculation

∠BOC = 360o – 90o – 110o

By subtraction

∠BOC = 360o – 200o

So we get

∠BOC = 160o

We know that

∠BOC = 2 × ∠BAC

It is given that ∠BOC = 160o

∠BAC = 160o/2

By division

∠BAC = 80o

Therefore, ∠BAC = 80o.

2. In the given figure, O is the centre of the circle and ∠AOB = 70o. Calculate the values of

(i) ∠OCA,

(ii) ∠OAC. Solution:

(i) We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠OCA

It is given that ∠AOB = 70o

We can write it as

∠OCA = 70o/2

By division

∠OCA = 35o

Therefore, ∠OCA = 35o.

(ii) From the figure we know that the radius is

OA = OC

We know that the base angles of an isosceles triangle are equal

∠OAC = ∠OCA

It is given that ∠OCA = 35o

So we get

∠OAC = 35o

Therefore, ∠OAC = 35o.

3. In the given figure, O is the centre of the circle. If ∠PBC = 25o and ∠APB = 110o, find the value of ∠ADB. Solution:

We know that ∠ACB = ∠PCB In △ PCB

Using the angle sum property

∠PCB + ∠BPC + ∠PBC = 180o

We know that ∠APB and ∠BPC are linear pair

By substituting the values

∠PCB + (180o – 110o) + 25o = 180o

On further calculation

∠PCB + 70o + 25o = 180o

∠PCB + 95o = 180o

By subtraction

∠PCB = 180o – 95o

So we get

∠PCB = 85o

We know that the angles in the same segment of a circle are equal

Therefore, the value of ∠ADB is 85o.

4. In the given figure, O is the centre of the circle. If ∠ABD = 35o and ∠BAC = 70o, find ∠ACB. Solution:

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

Using the angle sum property

By substituting the values

∠ADB + 90o + 35o = 180o

On further calculation

∠ADB = 180o – 90o – 35o

By subtraction

So we get

We know that the angles in the same segment of a circle are equal

So we get

∠ACB = 55o

Therefore, ∠ACB = 55o.

5. In the given figure, O is the centre of the circle. If ∠ACB = 50o, find ∠OAB. Solution:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

It is given that ∠ACB = 50o

By substituting

∠AOB = 2 × 50o

By multiplication

∠AOB = 100o …… (1)

Consider △ OAB

We know that the radius of the circle are equal

OA = OB

Base angles of an isosceles triangle are equal

So we get

∠OAB = ∠OBA ……. (2)

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180o

Using equations (1) and (2) we get

100o + 2 ∠OAB = 180o

By subtraction

2 ∠OAB = 180o – 100o

So we get

2 ∠OAB = 80o

By division

∠OAB = 40o

Therefore, ∠OAB = 40o.

6. In the given figure, ∠ABD = 54o and ∠BCD = 43o, calculate

(i) ∠ACD,

(iii) ∠BDA. Solution:

(i) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠ABD and ∠ACD are in the segment AD

∠ABD = ∠ACD = 54o

(ii) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠BAD and ∠BCD are in the segment BD

(iii) In △ ABD

Using the angle sum property

By substituting the values

43o + ∠ADB + 54o = 180o

On further calculation

∠ADB = 180o – 43o – 54o

By subtraction

So we get

It can be written as

∠BDA = 83o

7. In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If ∠CBD = 60o, calculate ∠CDE. Solution:

We know that the angles in the same segment of a circle are equal

From the figure we know that ∠CAD and ∠CBD are in the segment CD

An angle in a semi-circle is a right angle

So we get

Using the angle sum property

By substituting the values

∠ACD + 90o + 60o = 180o

On further calculation

∠ACD = 180o – 90o – 60o

By subtraction

∠ACD = 180o – 150o

So we get

∠ACD = 30o

We know that AC || DE and CD is a transversal

From the figure we know that ∠CDE and ∠ACD are alternate angles

So we get

∠CDE = ∠ACD = 30o

Therefore, ∠CDE = 30o.

8. In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ∠ABC = 25o, calculate ∠CED. Solution:

We know that ∠BCD and ∠ABC are alternate interior angles

∠BCD = ∠ABC = 25o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

∠BOD = 2 ∠BCD

It is given that ∠BCD = 25o

So we get

∠BOD = 2 (25o)

By multiplication

∠BOD = 50o

In the same way

∠AOC = 2 ∠ABC

So we get

∠AOC = 50o

From the figure we know that AB is a straight line passing through the centre

Using the angle sum property

∠AOC + ∠COD + ∠BOD = 180o

By substituting the values

50o + ∠COD + 50o = 180o

On further calculation

∠COD + 100o = 180o

By subtraction

∠COD = 180o – 100o

So we get

∠COD = 80o

We know that

∠CED = ½ ∠COD

So we get

∠CED = 80o/2

By division

∠CED = 40o

Therefore, ∠CED = 40o.

9. In the given figure, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80o and ∠CDE = 40o, find

(i) ∠DCE,

(ii) ∠ABC. Solution:

(i) From the figure we know that ∠CED = 90o

Consider △ CED

Using the angle sum property

∠CED + ∠EDC + ∠DCE = 180o

By substituting the values

90o + 40o + ∠DCE = 180o

On further calculation

∠DCE = 180o – 90o – 40o

By subtraction

∠DCE = 180o – 130o

So we get

∠DCE = 50o

(ii) We know that ∠AOC and ∠BOC form a linear pair

It can be written as

∠BOC = 180o – 80o

By subtraction

∠BOC = 100o

Using the angle sum property

∠ABC + ∠BOC + ∠DCE = 180o

By substituting the values

∠ABC + 100o + 50o = 180o

On further calculation

∠ABC = 180o – 100o – 50o

By subtraction

∠ABC = 180o – 150o

So we get

∠ABC = 30o

10. In the given figure, O is the centre of a circle, ∠AOB = 40o and ∠BDC = 100o, find ∠OBC. Solution:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

From the figure we know that

∠ACB = ∠DCB

It can be written as

∠AOB = 2 ∠DCB

We also know that

∠DCB = ½ ∠AOB

By substituting the values

∠DCB = 40o/2

By division

∠DCB = 20o

In △ DBC

Using the angle sum property

∠BDC + ∠DCB + ∠DBC = 180o

By substituting the values we get

100o + 20o + ∠DBC = 180o

On further calculation

∠DBC = 180o – 100o – 20o

By subtraction

∠DBC = 180o – 120o

So we get

∠DBC = 60o

From the figure we know that

∠OBC = ∠DBC = 60o

So we get

∠OBC = 60o

Therefore, ∠OBC = 60o.

11. In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If ∠OAB = 25o, calculate ∠EBC. Solution:

We know that OA and OB are the radius

Base angles of an isosceles triangle are equal

So we get

∠OBA = ∠OAB = 25o

Consider △ OAB

Using the angle sum property

∠OAB + ∠OBA + ∠AOB = 180o

By substituting the values

25o + 25o + ∠AOB = 180o

On further calculation

∠AOB = 180o – 25o – 25o

By subtraction

∠AOB = 180o – 50o

So we get

∠AOB = 130o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

∠AOB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠AOB

By substituting the values

∠ACB = 130/2

By division

∠ACB = 65o

So we get

∠ECB = 65o

In △ BEC

Using the angle sum property

∠EBC + ∠BEC + ∠ECB = 180o

By substituting the values

∠EBC + 90o + 65o = 180o

On further calculation

∠EBC = 180o – 90o – 65o

By subtraction

∠EBC = 180o – 155o

So we get

∠EBC = 25o

Therefore, ∠EBC = 25o.

12. In the given figure, O is the centre of a circle in which ∠OAB = 20o and ∠OCB = 55o. Find

(i) ∠BOC,

(ii) ∠AOC. Solution:

(i) We know that OB = OC which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBC = ∠OCB = 55o

In △ BOC

Using the angle sum property

∠BOC + ∠OCB + ∠OBC = 180o

By substituting the values

∠BOC + 55o + 55o = 180o

On further calculation

∠BOC = 180o – 55o – 55o

By subtraction

∠BOC = 180o – 110o

So we get

∠BOC = 70o

(ii) We know that OA = OB which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBA= ∠OAB = 20o

In △ AOB

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180o

By substituting the values

∠AOB + 20o + 20o = 180o

On further calculation

∠AOB = 180o – 20o – 20o

By subtraction

∠AOB = 180o – 40o

So we get

∠AOB = 140o

We know that

∠AOC = ∠AOB – ∠BOC

By substituting the values

∠AOC = 140o – 70o

So we get

∠AOC = 70o

13. In the given figure, O is the centre of the circle and ∠BCO = 30o. Find x and y. Solution:

From the figure we know that ∠AOD and ∠OEC form right angles

So we get

∠AOD = ∠OEC = 90o

We know that OD || BC and OC is a transversal

From the figure we know that ∠AOD and ∠OEC are corresponding angles

∠AOD = ∠OEC

We know that ∠DOC and ∠OCE are alternate angles

∠DOC = ∠OCE = 30o

Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

So we get

∠DOC = 2 ∠DBC

It can be written as

∠DBC = ½ ∠DOC

By substituting the values

∠DBC = 30/2

By division

y = ∠DBC = 15o

In the same way

∠ABD = ½ ∠AOD

By substituting the values

∠ABD = 90/2

By division

∠ABD = 45o

We know that

∠ABE = ∠ABC = ∠ABD + ∠DBC

So we get

∠ABE = ∠ABC = 45o + 15o

∠ABE = ∠ABC = 60o

Consider △ ABE

Using the angle sum property

∠BAE + ∠AEB + ∠ABE = 180o

By substituting the values

x + 90o + 60o = 180o

On further calculation

x = 180o – 90o – 60o

By subtraction

x = 180o – 150o

So we get

x = 30o

Therefore, the value of x is 30o and y is 15o.

14. In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB. Solution:

Join the points AC It is given that BD = OD

We know that the radii of same circle are equal

OD = OB

It can be written as

BD = OD = OB

Consider △ ODB as equilateral triangle

We know that

∠ODB = 60o

The altitude of an equilateral triangle bisects the vertical angle

So we get

∠BDE = ∠ODE = ½ ∠ODB

By substituting the values

∠BDE = ∠ODE = 60/2

By division

∠BDE = ∠ODE = 30o

We know that the angles in the same segment are equal

∠CAB = ∠BDE = 30o

Therefore, ∠CAB = 30o.

15. In the given figure, PQ is a diameter of a circle with centre O. If ∠PQR = 65o, ∠SPR = 40o and ∠PQM = 50o, find ∠QPR, ∠QPM and ∠PRS. Solution:

In △ PQR

We know that PQ is the diameter

So we get

∠PRQ = 90o as the angle in a semicircle is a right angle

Using the angle sum property

∠QPR + ∠PRQ + ∠PQR = 180o

By substituting the values

∠QPR + 90o + 65o = 180o

On further calculation

∠QPR = 180o – 90o – 65o

By subtraction

∠QPR = 180o – 155o

So we get

∠QPR = 25o

In △ PQM

We know that PQ is the diameter

So we get

∠PMQ = 90o as the angle in a semicircle is a right angle

Using the angle sum property

∠QPM + ∠PMQ + ∠PQM = 180o

By substituting the values

∠QPM + 90o + 50o = 180o

On further calculation

∠QPM = 180o – 90o – 50o

By subtraction

∠QPM = 180o – 140o

So we get

∠QPM = 40o

We know that

∠QPS + ∠SRQ = 180o

It can be written as

∠QPR + ∠RPS + ∠PRQ + ∠PRS = 180o

By substituting the values

25o + 40o + 90o + ∠PRS = 180o

On further calculation

∠PRS = 180o – 25o – 40o – 90o

By subtraction

∠PRS = 180o – 155o

So we get

∠PRS = 25o

16. In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line. If ∠APB = 150o and ∠BQD = xo, find the value of x. Solution:

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠APB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠APB

By substituting the values

∠ACB = 150/2

So we get

∠ACB = 75o

We know that ACD is a straight line

It can be written as

∠ACB + ∠DCB = 180o

By substituting the values

75o + ∠DCB = 180o

On further calculation

∠DCB = 180o – 75o

By subtraction

∠DCB = 105o

We know that

∠DCB = ½ × reflex ∠BQD

By substituting the values

105o = ½ × (360o – x)

On further calculation

210o = 36o – x

By subtraction

x = 150o

Therefore, the value of x is 150o.

17. In the given figure, ∠BAC = 30o. Show that BC is equal to the radius of the circumcircle of △ ABC whose centre is O. Solution:

Join the lines OB and OC

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

∠BOC = 2 ∠BAC

It is given that ∠BAC = 30o

∠BOC = 2 × 30o

By multiplication

∠BOC = 60o

In △ BOC

We know that the radii are equal

OB = OC

The base angles of an isosceles triangle are equal

∠OBC = ∠OCB

Consider △ BOC

Using the angle sum property

∠BOC + ∠OBC + ∠OCB = 180o

By substituting the values we get

60o + ∠OCB + ∠OCB = 180o

So we get

2 ∠OCB = 180o – 60o

By subtraction

2 ∠OCB = 120o

By division

∠OBC = 60o

We get ∠OBC = ∠OCB = ∠BOC = 60o

We know that △ BOC is an equilateral triangle

So we get OB = OC = BC

Therefore, it is proved that BC is equal to the radius of the circumcircle of △ ABC whose centre is O.

18. In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E. Prove that ∠AEC = ½ (angle subtended by arc CXA. At the centre + angle subtended by arc DYB at the centre). Solution:

Construct the lines AC and BC The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

So we get

∠AOC = 2 ∠ABC …… (1)

In the same way

∠BOD = 2 ∠BCD …… (2)

∠AOC + ∠BOD = 2 ∠ABC + 2 ∠BCD

By taking 2 as common

∠AOC + ∠BOD = 2 (∠ABC + ∠BCD)

It can be written as

∠AOC + ∠BOD = 2 (∠EBC + ∠BCE)

So we get

∠AOC + ∠BOD = 2 (180o – ∠CEB)

We can write it as

∠AOC + ∠BOD = 2 (180o – (180o – ∠AEC))

We get

∠AOC + ∠BOD = 2 ∠AEC

Dividing the equation by 2

∠AEC = ½ (∠AOC + ∠BOD)

∠AEC = ½ (angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre)

Therefore, it is proved that ∠AEC = ½ (angle subtended by arc CXA. At the centre + angle subtended by arc DYB at the centre).

### Access other exercise solutions of Class 9 Maths Chapter 12: Circles

Exercise 12A Solutions 23 Questions

Exercise 12C Solutions 26 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 12 – Circles Exercise 12B

RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Exercise 12B contains the theorems which are based on the results of angles which are subtended by arcs and problems regarding them.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 12: Circles Exercise 12B

• Studying concepts in Mathematics is made simple with the help of RS Aggarwal Solutions in PDF format.
• The solutions are in a stepwise manner which improves problem solving abilities among students.
• Several examples before each exercise mainly help students gain conceptual knowledge before solving the problems.
• Theorems are explained in brief as each of them carry more marks in the board exam according to the CBSE syllabus.