Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles Exercise 24(A) provides comprehensive answers on each and every question discussed under this exercise. The solutions are designed with the purpose of improving students’ knowledge in the respective subject. Those who aim to obtain skills in problem solving are advised to practice Selina Solutions frequently. The solutions can be referred by the students to score more marks in the annual exam. For a better hold on these concepts, students can download Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles Exercise 24(A) free PDF, from the links provided below.

## Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles Exercise 24(A) Download PDF

### Access another exercise of Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles

### Access Selina Solutions Concise Mathematics Class 6 Chapter 24: Angles Exercise 24(A)

**1. For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.**

**(iv) Name the angles marked by letters a, b, c, x and y.**

**Solution:**

(i) In the given figure,

Vertex = O

Arms = OA and OB

Angle = ∠AOB or ∠BOA or ∠O

(ii) In the given figure,

Vertex = Q

Arms = QP and QR

Angle = ∠PQR or ∠RQP or ∠Q

(iii) In the given figure,

Vertex = M

Arms = MN and ML

Angle = ∠LMN or ∠NML or ∠M

(iv) a = ∠AOE

b = ∠AOB

c = ∠BOC

d = ∠COD

e = ∠DOE

**2. Name the points:**

**(i) in the interior of the angle PQR,**

**(ii) in the exterior of the angle PQR.**

**Solution:**

(i) The points in the interior of the angle = a, b and x

(ii) The points in the exterior of the angle = d, m, n, s and t

**3. In the given figure, figure out the number of angles formed within the arms OA and OE.**

**Solution:**

In the given figure, the angles within the arms OA and OE are as follows:

(i) ∠AOE

(ii) ∠AOD

(iii) ∠AOC

(iv) ∠AOB

(v) ∠BOC

(vi) ∠BOD

(vii) ∠BOE

(viii) ∠COD

(ix) ∠COE and

(x) ∠DOE

**4. Add:**

**(i) 29 ^{0 }16’23” and 8^{0 }27’12”**

**(ii) 9 ^{0} 45’56” and 73^{0} 8’ 15”**

**(iii) 56 ^{0} 38’ and 27^{0} 42’ 30”**

**(iv) 47 ^{0} and 61^{0} 17’4”**

**Solution:**

(i) 29^{0 }16’23” + 8^{0 }27’12”

29^{0 }16’23”

8^{0 }27’12” +

_____________

37^{0} 43’ 35”

_____________

Hence, addition of 29^{0 }16’23” and 8^{0 }27’12” = 37^{0} 43’ 35”

(ii) 9^{0} 45’56” + 73^{0} 8’ 15”

9^{0} 45’56”

73^{0} 8’ 15” +

____________

82^{0} 53’ 71”

____________

Hence, addition of 9^{0} 45’56” and 73^{0} 8’ 15” = 82^{0} 53’ 71”

(iii) 56^{0} 38’ + 27^{0} 42’ 30”

56^{0} 38’

27^{0} 42’ 30” +

_____________

83^{0} 80’ 30”

_____________

Hence, addition of 56^{0} 38’ and 27^{0} 42’ 30” = 83^{0} 80’ 30”

(iv) 47^{0} + 61^{0} 17’4”

47^{0}

61^{0} 17’4” +

____________

108^{0} 17’ 4”

_____________

Hence, addition of 47^{0} and 61^{0} 17’4” = 108^{0} 17’ 4”

**5. In the figure, given below name:**

**(i) three pairs of adjacent angles**

**(ii) two acute angles**

**(iii) two obtuse angles**

**(iv) two reflex angles**

**Solution:**

(i) In the given figure, the three pairs of adjacent angles are as follows:

∠AOB and ∠BOC ,

∠BOC and ∠COD

∠COD and ∠DOA

(ii) In the given figure, the two acute angles are

∠AOB and ∠AOD

(iii) In the given figure, the two obtuse angles are

∠BOC and ∠COD

(iv) In the given figure, the two reflex angles are

∠AOB and ∠COB

**6. In the given figure; PQR is a straight line. If:**

**(i) ∠SQR = 75 ^{0}; find ∠PQS.**

**(ii) ∠PQS = 110 ^{0}; find ∠RQS**

**Solution:**

(i) Given PQR is a straight line

In the given figure,

∠PQS + ∠SQR = 180^{0} {linear pair of angles}

∠PQ S + 75 = 180

On further calculation, we get,

∠PQS = 180 – 75

∠PQS = 105^{0}

(ii) Given PQR is a straight line

∠PQS + ∠RQS = 180^{0}

110^{0} + ∠RQS = 180^{0}

On further calculation, we get,

∠RQS = 180^{0} – 110^{0}

∠RQS = 70^{0}

**7. In the given figure; AOC is a straight line. If angle AOB = 50 ^{0}, angle AOE = 90^{0} and angle COD = 25^{0}; find the measure of:**

**(i) angle BOC**

**(ii) angle EOD**

**(iii) obtuse angle BOD**

**(iv) reflex angle BOD**

**(v) reflex angle COE**

**Solution:**

(i) Given

AOC is a straight line

∠AOB = 50^{0}

∠AOE = 90^{0}

∠COD = 25^{0}

To find the measure of ∠BOC

∠AOB + ∠BOC = 180^{0} (Linear pairs of angle)

50^{0} + ∠BOC = 180^{0}

On further calculation, we get,

∠BOC = 180^{0} – 50^{0}

∠BOC = 130^{0}

(ii) Given

AOC is a straight line

∠AOB = 50^{0}

∠AOE = 90^{0}

∠COD = 25^{0}

To find the measure ∠EOD

∠EOD + ∠COD = 90^{0} (Since ∠AOE = 90^{0})

∠EOD + 25^{0} = 90^{0}

∠EOD = 90^{0} – 25^{0}

We get,

∠EOD = 65^{0}

(iii) Given

AOC is a straight line

∠AOB = 50^{0}

∠AOE = 90^{0}

∠COD = 25^{0}

To find the measure of obtuse angle BOD,

∠BOD = ∠BOC + ∠COD

Substituting the value of ∠BOC and ∠COD, we get,

∠BOD = 130^{0} + 25^{0}

We get,

∠BOD = 155^{0}

(iv) Given

AOC is a straight line

∠AOB = 50^{0}

∠AOE = 90^{0}

∠COD = 25^{0}

To find the measure of reflex angel BOD

∠BOD = 360^{0} – ∠BOD

= 360^{0} – 155^{0}

We get,

∠BOD = 205^{0}

(iv) Given

AOC is a straight line

∠AOB = 50^{0}

∠AOE = 90^{0}

∠COD = 25^{0}

To find the measure of reflex angle COE

∠COE = 360^{0} – ∠COE

= 360^{0} – (∠COD + ∠EOD)

= 360^{0} – (25^{0} + 65^{0})

We get,

= 360^{0} – 90^{0}

∠COE = 270^{0}

**8. In the given figure if:**

**(i) a = 130 ^{0}; find b**

**(ii) b = 200; find a**

**(iii) a = 5 / 3 right angle, find b**

**Solution:**

(i) From figure,

a + b = 360^{0}

Substitute the value of a in above equation

130^{0} + b = 360^{0}

b = 360^{0} – 130^{0}

We get,

b = 230^{0}

(ii) From figure,

a + b = 360^{0}

a + 200^{0} = 360^{0}

a = 360^{0} – 200^{0}

We get,

a = 160^{0}

(iii) From figure,

a = 5 / 3 right angle

= 5 / 3 × 90^{0}

We get,

a = 150^{0}

Now,

a + b = 360^{0}

Substitute the value of a in above equation

150^{0} + b = 360^{0}

b = 360^{0} – 150^{0}

b = 210^{0}

**9. In the given diagram, ABC is a straight line**

**(i) If x = 53 ^{0}, find y**

**(ii) If y = **

** right angles; find x.**

**Solution:**

(i) From the figure,

Given that ABC is a straight line

Hence,

∠ABD + ∠DBC = 180^{0} {Linear pair of angles}

x + y = 180^{0}

53^{0} + y = 180^{0}

y = 180^{0} – 53^{0}

y = 127^{0}

(ii) From given figure,

x + y = 180^{0}

x +

right angles = 180^{0}

x + 3 / 2 × 90^{0} = 180^{0}

On further calculation, we get

x + 135^{0} = 180^{0}

x = 180^{0} – 135^{0}

x = 45^{0}

**10. In the given figure, AOB is a straight line. Find the value of x and also answer each of the following:**

**(i) ∠AOP = …….**

**(ii) ∠BOP = …….**

**(iii) which angle is obtuse?**

**(iv) which angle is acute?**

**Solution:**

Given from figure,

∠AOP = x + 30^{0}

∠BOP = x – 30^{0}

Now,

∠AOP + ∠BOP = 180^{0} (Since ∠AOB is a straight line)

(x + 30^{0})+ (x – 30^{0}) = 180^{0}

2x = 180^{0}

We get,

x = 90^{0}

(i) ∠AOP = x + 30^{0}

= 90^{0} + 30^{0}

= 120^{0}

(ii) ∠BOP = x – 30^{0}

= 90^{0} – 30^{0}

We get,

= 60^{0}

(iii) The obtuse angle is ∠AOP

(iv) The acute angle is ∠BOP

**11.In the given figure, PQR is a straight line. Find x. Then complete the following:**

**(i) ∠AQB = ……………**

**(ii) ∠BQP = ……………**

**(iii) ∠AQR = …………….**

**Solution:**

Given

PQR is a straight line

∠AQP = x + 20^{0}

∠AQB = 2x + 10^{0}

∠BQR = x – 10^{0}

Since PQR is a straight line

∠AQP + ∠AQB + ∠BQR = 180^{0}

(x + 20^{0}) + (2x + 10^{0}) + (x – 10^{0}) = 180^{0}

We get,

4x + 20^{0} = 180^{0}

4x = 180^{0} – 20^{0}

4x = 160^{0}

x = 160^{0} / 4

We get,

x = 40^{0}

(i) ∠AQB = 2x + 10^{0}

= 2 × 40^{0} + 10^{0}

= 80^{0} + 10^{0}

= 90^{0}

Similarly,

∠AQP = x + 20^{0}

∠AQP = 40^{0} + 20^{0}

∠AQP = 60^{0}

∠BQR = x – 10^{0}

∠BQR = 40^{0} – 10^{0}

∠BQR = 30^{0}

(ii) ∠BQP = ∠AQP + ∠AQB

= 60^{0} + 90^{0}

= 150^{0}

(iii) ∠AQR = ∠AQB + ∠BQR

= 90^{0} + 30^{0}

= 120^{0}

**12. In the given figure, lines AB and CD intersect at point O.**

**(i) Find the value of ∠a.**

**(ii) Name all the pairs of vertically opposite angles.**

**(iii) Name all the pairs of adjacent angles**

**(iv) Name all the reflex angles formed and write the measure of each.**

**Solution:**

Given

AB and CD intersect each other at point C

∠AOC = 68^{0}

(i) Here, AOB is a line

Hence,

∠AOC + ∠BOC = 180^{0} (Linear pairs of angles)

68^{0} + a = 180^{0}

a = 180^{0} – 68^{0}

a = 112^{0}

(ii) The pairs of vertically opposite angles are,

∠AOC and ∠BOD and ∠BOC and ∠AOD

(iii) The pairs of adjacent angles are,

∠AOC and ∠BOC, ∠BOC and ∠BOD, ∠BOD and ∠DOA, ∠DOA and ∠AOC

(iv) The reflex angles in the given figure are,

∠BOC and ∠DOA,

Reflex angle BOC = 180^{0} + 68^{0}

= 248^{0}

Reflex angle DOA = 180^{0} + 68^{0}

= 248^{0}

**13. In the given figure:**

**(i) If ∠AOB = 45 ^{0}, ∠BOC = 30^{0} and ∠AOD = 110^{0};**

**Find: angles COD and BOD**

**(ii) If ∠BOC = ∠DOC = 34 ^{0} and ∠AOD = 120^{0};**

**Find: angle AOB and angle AOC**

**(iii) If ∠AOB = ∠BOC = ∠COD = 38 ^{0}**

**Find: reflex angle AOC and reflex angle AOD**

**Solution:**

(i) ∠COD = ∠AOD – ∠AOC

= ∠AOD – (∠AOB + ∠BOC)

= 110^{0} – (45^{0} + 30^{0})

We get,

= 110^{0} – 75^{0}

= 35^{0}

Hence,

∠COD = 35^{0}

∠BOD = ∠BOC + ∠COD

= 30^{0} + 35^{0}

= 65^{0}

Hence,

∠BOD = 65^{0}

(ii) ∠AOB = ∠AOD – ∠BOD

= ∠AOD – (∠BOC + ∠COD)

= 120^{0} – (34^{0} + 34^{0})

We get,

= 120^{0} – 68^{0}

= 52^{0}

Hence,

∠AOB = 52^{0}

∠AOC = ∠AOB + ∠BOC

= 52^{0} + 34^{0}

= 86^{0}

Hence,

∠AOC = 86^{0}

(iii) Reflex angel AOC = 360^{0} – ∠AOC

= 360^{0} – (∠AOB + ∠BOC)

= 360^{0} – (38^{0} + 38^{0})

= 360^{0} – 76^{0}

We get,

= 284^{0}

Hence,

∠AOC = 284^{0}

Reflex angle AOD = 360^{0} – ∠AOD

= 360^{0} – (∠AOB + ∠BOC + ∠COD)

= 360^{0} – (38^{0} + 38^{0} + 38^{0})

= 360^{0} – 114^{0}

We get,

= 246^{0}

Hence,

∠AOC = 246^{0}