Mathematics Solutions Solutions for Class 7 Maths Chapter 4 Angles And Pairs Of Angles are provided here with simple step-by-step explanations. These solutions for Angles And Pairs Of Angles are extremely popular among Class 7 students for Maths Angles And Pairs Of Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 25:

#### Question 1:

Observe the figure and complete the table for ∠AWB.

Points in the interior | |

Points in the exterior | |

Points on the arms of the angles |

#### Answer:

Points in the interior | R, C, N, X |

Points in the exterior | T, U, Q, V, Y |

Points on the arms of the angles | A, W, G, B |

#### Page No 25:

#### Question 2:

Name the pairs of adjacent angles in the figures below.

#### Answer:

Two angles which have a common vertex, a common arm and separate interiors are said to be adjacent angles.

The pairs of adjacent angles are given below:

∠ANB and ∠BNC,

∠BNC and ∠ANC,

∠ANC and ∠ANB,

∠PQR and ∠PQT

#### Page No 25:

#### Question 3:

#### Answer:

Two angles which have a common vertex, a common arm and separate interiors are said to be adjacent angles

(i)

In ∠PMQ and ∠RMQ, M is the common vertex and MQ is the common arm.

Therefore, ∠PMQ and ∠RMQ are adjacent angles.

(ii)

The angles ∠RMQ and ∠SMR have a common vertex M, but don't have common arm.

Therefore, ∠RMQ and ∠SMR are not adjacent angles.

(iii)

The angles ∠RMS and ∠RMT have a common vertex M, but don't have common arm.

Therefore, ∠RMS and ∠RMT are not adjacent angles.

(iv)

In ∠SMT and ∠RMS, M is the common vertex and SM is the common arm.

Therefore, ∠SMT and ∠RMS are adjacent angles.

#### Page No 26:

#### Question 1:

^{°}(ii) 63

^{°}(iii) 45

^{°}(iv) 55

^{°}(v) 20

^{°}(vi) 90

^{°}(vii)

*x*

^{°}

#### Answer:

(i)

Let the measure of the complementary angle be *a*.

40 + *a* = 90

∴ *a* = 50°

Hence, the measure of the complement of an angle of measure 40° is 50°

(ii)

Let the measure of the complementary angle be *a*.

63 + *a* = 90

∴ *a* = 27°

Hence, the measure of the complement of an angle of measure 63° is 27°

(iii)

Let the measure of the complementary angle be *a*.

45 + *a* = 90

∴ *a* = 45°

Hence, the measure of the complement of an angle of measure 45° is 45°

(iv)

Let the measure of the complementary angle be *a*.

55 + *a* = 90

∴ *a* = 35°

Hence, the measure of the complement of an angle of measure 55° is 35°

(v)

Let the measure of the complementary angle be *a*.

20 + *a* = 90

∴ *a* = 70°

Hence, the measure of the complement of an angle of measure 20° is 70°

(vi)

Let the measure of the complementary angle be *a*.

90 + *a* = 90

∴ *a* = 0°

Hence, the measure of the complement of an angle of measure 00° is 0°

(vii)

Let the measure of the complementary angle be *a*.

*x* + *a* = 90

∴ *a* = (90 − *x*)°

Hence, the measure of the complement of an angle of measure *x*° is (90 − *x*)°

#### Page No 26:

#### Question 2:

(y $-$ 20)^{° }and (y + 30)^{°} are the measures of complementary angles. Find the measure of each angle.

#### Answer:

Sum of two complementary angles is 90^{°}

∴ (*y* $-$ 20)^{°} + (*y* + 30)^{°} = 90^{°}

⇒ *y* $-$ 20 + *y* + 30 = 90

⇒ 2*y* + 10 = 90

⇒ 2*y* = 80

⇒ *y* = 40

Hence, the measure of the two angles are 20^{°} and 70^{°}.

#### Page No 27:

#### Question 1:

^{° }(ii) 85

^{° }(iii) 120

^{°}(iv) 37

^{° }(v) 108

^{° }(vi) 0

^{° }(vii)

*a*

^{°}

#### Answer:

(i)

Let the measure of the supplementary angle be *a*.

15 + *a* = 180

∴ *a* = 165°

Hence, the measure of the supplement of an angle of measure 15° is 165°.

(ii)

Let the measure of the supplementary angle be *a*.

85 + *a* = 180

∴ *a* = 95°

Hence, the measure of the supplement of an angle of measure 85° is 95°.

(iii)

Let the measure of the supplementary angle be *a*.

120 + *a* = 180

∴ *a* = 60°

Hence, the measure of the supplement of an angle of measure 120° is 60°.

(iv)

Let the measure of the supplementary angle be *a*.

37 + *a* = 180

∴ *a* = 143°

Hence, the measure of the supplement of an angle of measure 37° is 143°.

(v)

Let the measure of the supplementary angle be *a*.

108 + *a* = 180

∴ *a* = 72°

Hence, the measure of the supplement of an angle of measure 108° is 72°.

(vi)

Let the measure of the supplementary angle be *a*.

0 + *a* = 180

∴ *a* = 180°

Hence, the measure of the supplement of an angle of measure 0° is 180°.

(vii)

Let the measure of the supplementary angle be *x*.

*a* + *x* = 180

∴ *x* = (180 − *a*)°

Hence, the measure of the supplement of an angle of measure *a*° is (180 − *a*)°.

#### Page No 27:

#### Question 2:

*m*∠B = 60

^{°}

*m*∠N = 30

^{° }

*m*∠Y = 90

^{° }

*m*∠J = 150

^{°}

*m*∠D = 75

^{° }

*m*∠E = 0

^{°}

*m*∠F = 15

^{°}

*m*∠G = 120

^{°}

#### Answer:

If the sum of the measures of two angles is 90° they are known as complementary angles.

Hence,the pairs of complementary angles are ∠B and ∠N, ∠D and ∠F, ∠Y and ∠E.

If the sum of the measures of two angles is 180° they are known as supplementary angles.

Hence, the pairs of supplementary angles are ∠B and ∠G, ∠N and ∠J.

#### Page No 27:

#### Question 3:

In ∆XYZ, *m*∠Y = 90^{°}. What kind of a pair do ∠X and ∠Z make?

#### Answer:

In ∆XYZ,

∠X + ∠Y + ∠Z = 180^{°} (Angle Sum property of triangle

⇒ ∠X + 90^{°} + ∠Z = 180^{°}

⇒ ∠X + ∠Z = 90^{°}

Since, the sum of the measure of the two angles is 90^{°}.

Hence, ∠X and ∠Z are complementary angles.

#### Page No 27:

#### Question 4:

^{°}. Find the measures of the two angles.

#### Answer:

Let the measure of the first angle *a*.

Then, the measure of the other angle *a + *40°

Now, *a* + *a* + 40 = 90

⇒ 2*a* = 50

⇒ *a* = 25°

Hence, the measure of the two angles are 25° and 65°.

#### Page No 27:

#### Question 5:

$\square $ PTNM is a rectangle. Write the names of the pairs of supplementary angles.

#### Answer:

If the sum of the measures of two angles is 180° they are known as supplementary angles.

The measure of all the angles of a rectangle is 90°.

Hence, the pairs of supplementary angles are ∠P and ∠M, ∠T and ∠N, ∠P and ∠T, ∠M and ∠N, ∠P and ∠N, ∠M and ∠T.

#### Page No 27:

#### Question 6:

*m*∠A = 70

^{°}, what is the measure of the supplement of the complement of ∠A?

#### Answer:

Let the measure of the complementary angle be *a*.

70 + *a* = 90

∴ *a* = 20°

Let the measure of the supplementary angle of 20° be *x*.

20 + *x* = 180

∴ *x* = 160°

Hence, the measure of the supplement of the complement of ∠A is 160°.

#### Page No 27:

#### Question 7:

*m*∠B = (x + 20)

^{°}, then what would be

*m*∠A?

#### Answer:

Let the measure of the supplementary angle of ∠B be *a*.

(*x* + 20)^{°} + *a* = 180

∴ *a* = (160 − *x*)°

Hence, the measure of ∠A is (160 − *x*)°.

#### Page No 28:

#### Question 1:

#### Answer:

Two rays which have a common origin and form a straight line are said to be opposite rays.

Hence, the pairs of opposite rays are ray PL & ray PM and ray PN & ray PT.

#### Page No 28:

#### Question 2:

Are the ray PM and PT opposite rays? Give reasons for your answer.

#### Answer:

Ray PM and PT are not opposite rays because they do not form a straight line.

#### Page No 29:

#### Question 1:

#### Answer:

(ii)

The sum of the measures of the angles in a linear pair is 180°.

Therefore, angles in a linear pair are always supplementary.

(iii)

(iv)

(v)

(vi)

If the sum of the measures of two angles is 180° they are known as supplementary angles.

The sum of the measures of the angles in a linear pair is 180°.

Therefore, angles in a linear pair are always supplementary.

#### Page No 30:

#### Question 1:

*m*∠APD = 47

^{°}.Find the measures of ∠APB, ∠BPC, ∠CPD.

#### Answer:

In the given figure,

∠DPA + ∠APB = 180^{∘} (Linear Pair angles)

⇒ 47^{∘} + ∠APB = 180^{∘}

⇒ ∠APB = 133^{∘}

Now,

∠APD = ∠BPC = 47^{∘} (Vertically opposite angles)

∠APB = ∠DPC = 133^{∘} (Vertically opposite angles)

Hence, the measures of ∠APB, ∠BPC, ∠CPD are 133^{∘}, 47^{∘} and 133^{∘} respectively.

#### Page No 30:

#### Question 2:

Lines PQ and RS intersect at point M. *m*∠PMR = *x*^{°} What are the measures of ∠PMS, ∠SMQ and ∠QMR?

#### Answer:

In the given figure,

∠RMP + ∠PMS = 180^{∘} (Linear Pair angles)

⇒ *x*^{∘} + ∠PMS = 180^{∘}

⇒ ∠PMS = (180 − *x*)^{∘}

Now,

∠PMR = ∠SMQ = *x*^{∘} (Vertically opposite angles)

∠PMS = ∠RMQ = (180 − *x*)^{∘} (Vertically opposite angles)

Hence, the measures of ∠PMS, ∠SMQ and ∠QMR are (180 − *x*)^{∘}, *x*^{∘} and (180 − *x*)^{∘} respectively.

#### Page No 33:

#### Question 1:

∠ACD is an exterior angle of $\u2206$ABC.The measures of ∠A and ∠B are equal. If *m*∠ACD = 140^{°}, find the measures of the angles ∠A and ∠B.

#### Answer:

∠A + ∠B = ∠ACD (Exterior angle property)

⇒ 2∠A = 140^{∘} (∵∠A = ∠B)

⇒ ∠A = 70^{∘}

Hence, the measures of ∠A and ∠B are 70^{∘} and 70^{∘} respectively.

#### Page No 33:

#### Question 2:

Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

#### Answer:

In the given figure,

∠BOC = ∠FOE = 4*y* (Vertically opposite angles)

∠EOD = ∠AOB = 8*y* (Vertically opposite angles)

∠AOF = ∠COD = 6*y* (Vertically opposite angles)

Now, ∠AOB + ∠BOC + ∠COD = 180^{∘} (Linear Pair angles)

⇒ 8*y* + 4*y* + 6*y* = 180^{∘}

⇒ 18*y *=* *180^{∘}

⇒ *y *=* *10^{∘}

Therefore,

∠BOC = 4*y*

= 40^{∘}

∠EOD = 8*y*

= 80^{∘}

∠AOF = 6*y*

= 60^{∘}

Hence, the measures of ∠BOC, ∠EOD, ∠AOF are 40^{∘}, 80^{∘} and 80^{∘} respectively.

#### Page No 33:

#### Question 3:

(3x$-$17)

^{°}and (8x + 10)

^{° }respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

#### Answer:

Given:

∠ACB = (3*x *$-$ 17)^{∘}

∠ACD = (8*x* + 10)^{∘}

Now, ∠ACB + ∠ACD = 180^{∘} (Linear Pair angles)

⇒ 3*x *− 17^{ }+ 8*x* + 10 = 180

⇒ 11*x *=* *187

⇒ *x *=* *17

Therefore,

∠ACB = (3*x *$-$ 17)^{∘}

= (51* *$-$ 17)^{∘}

= 34^{∘}

∠ACD = (8*x* + 10)^{∘}

= (136* *+ 10)^{∘}

= 146^{∘}

Now, ∠A + ∠B = ∠ACD (Exterior angle property)

⇒ 2∠A = 146^{∘} (∵∠A = ∠B)

⇒ ∠A = 73^{∘}

Hence, the measures of ∠ACB, ∠ACD, ∠A and ∠B are 146^{∘}, 34^{∘}, 73^{∘} and 73^{∘} respectively.

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