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.

#### 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

 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

#### Question 2:

Name the pairs of adjacent angles in the figures below.

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

#### Question 3:

Are the following pairs adjacent angles? If not, state the reason.
(i) ∠PMQ and ∠RMQ        (ii) ∠RMQ and ∠SMR
(iii) ∠RMS and ∠RMT      (iv) ∠SMT and ∠RMS

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.

#### Question 1:

The measures of some angles are given below. Write the measures of their complementaryangles.
(i) 40°   (ii) 63° (iii) 45°(iv) 55° (v) 20° (vi) 90° (vii) x°

(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.
xa = 90
∴ a = (90 − x
Hence, the measure of the complement of an angle of measure x° is (90 − x

#### Question 2:

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

Sum of two complementary angles is 90°
∴ (y $-$ 20)° + (y + 30)° = 90°
⇒ y $-$ 20 + y + 30 = 90
2y + 10 = 90
⇒ 2y = 80
⇒ y = 40
Hence, the measure of the two angles are 20° and 70°.

#### Question 1:

Write the measures of the supplements of the angles given below.
(i) 15° (ii) 85° (iii) 120° (iv) 37° (v) 108° (vi) 0° (vii) a°

(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)°.

#### Question 2:

The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.
m∠B = 60° m∠N = 30° m∠Y = 90°  m∠J = 150°
m∠D = 75° m∠E = 0° m∠F = 15° m∠G = 120°

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.

#### Question 3:

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

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.

#### Question 4:

The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles.

Let the measure of the first angle a.
Then, the measure of the other angle a + 40°
Now, a + a + 40 = 90
⇒ 2a = 50
⇒ a = 25°
Hence, the measure of the two angles are 25° and 65°.

#### Question 5:

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

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.

#### Question 6:

If m∠A = 70°, what is the measure of the supplement of the complement of ∠A?

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°.

#### Question 7:

If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A?

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​)°.

#### Question 1:

Name the pairs of opposite rays in the figure alongside.

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.

#### Question 2:

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

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

#### Question 1:

Draw the pairs of angles as described below.If that is not possible, say why.
(i) Complementary angles that are not adjacent.
(ii)   Angles in a linear pair which are not supplementary.
(iii) Complementary angles that do not form a linear pair.
(iv)  Adjacent angles which are not in a linear pair.
(v) Angles which are neither complementary nor adjacent.
(vi) Angles in a linear pair which are complementary.

(i)

(ii)
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.

(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.

#### Question 1:

Lines AC and BD intersect at point P. m∠APD = 47° .Find the measures of ∠APB, ∠BPC, ∠CPD.

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.

#### Question 2:

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

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.

#### Question 1:

∠ACD is an exterior angle of $∆$ABC.The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

∠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.

#### Question 2:

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

In the given figure,
∠BOC = ∠FOE = 4y           (Vertically opposite angles)
∠EOD = ∠AOB = 8y           (Vertically opposite angles)
∠AOF = ∠COD = 6y           (Vertically opposite angles)
Now, ∠AOB + ∠BOC + ∠COD = 180         (Linear Pair angles)
⇒ 8y + 4y + 6y = 180
⇒ 18= 180
⇒ = 10
Therefore,
∠BOC = 4y
= 40
∠EOD = 8y
= 80
​∠AOF = 6y
= 60
Hence, the measures of ∠BOC, ∠EOD, ∠AOF are 4080 and 80 respectively.

#### Question 3:

In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are
(3x$-$17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

Given:
∠ACB = (3$-$ 17)
∠ACD = (8x + 10)
Now, ∠ACB + ∠ACD = 180         (Linear Pair angles)
⇒ 3− 17 8x + 10 = 180
⇒ 11x = 187
= 17
Therefore,
∠ACB = (3$-$ 17)
= (51 $-$ 17)
= 34
​∠ACD = (8x + 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 1463473 and 73 respectively.

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