## Maharashtra State Board Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles Practice Set 17

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°

Solution:

i. Let the measure of the supplementary angle be x°.

∴ 15 + x = 180

∴ 15 + x – 15 = 180 – 15

….(Subtracting 15 from both sides)

∴ x = 165

∴ The measures of the supplement of an angle of 15° is 165°.

ii. Let the measure of the supplementary angle be x°.

∴ 85 + x = 180

∴ 85 + x – 85 = 180 – 85

….(Subtracting 85 from both sides)

∴ x = 95

∴ The measures of the supplement of an angle of 85° is 95°.

iii. Let the measure of the supplementary angle be x°.

∴ 120 + x = 180

∴ 120 + x – 120 = 180 – 120

….(Subtracting 120 from both sides)

∴ x = 60

∴ The measures of the supplement of an angle of 120° is 60°.

iv. Let the measure of the supplementary angle be x°.

∴ 37 + x = 180

∴ 37 + x – 37 = 180 – 37

….(Subtracting 37 from both sides)

∴ x = 143

∴ The measures of the supplement of an angle of 37° is 143°.

v. Let the measure of the supplementary angle be x°.

∴ 108 + x = 180

∴ 108 + x – 108 = 180 – 108

….(Subtracting 108 from both sides)

∴ x = 72

∴ The measures of the supplement of an angle of 108° is 72°.

vi. Let the measure of the supplementary angle be x°.

∴0 + x = 180

∴ x = 180

∴ The measures of the supplement of an angle of 0° is 180°.

vii. Let the measure of the supplementary angle be x°.

∴ a + x = 180

∴ a + x – a = 180 – a

….(Subtracting a from both sides) x = (180 – a)

∴ The measures of the supplement of an angle of 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°

Solution:

i. m∠B + m∠N = 60° + 30°

= 90°

∴∠B and ∠N are a pair of complementary angles.

ii. m∠Y + m∠E = 90° + 0°

= 90°

∴∠Y and ∠E are a pair of complementary angles.

iii. m∠D + m∠F = 75° + 15°

= 90°

∴∠D and ∠F are a pair of complementary angles.

iv. m∠B + m∠G = 60° + 120°

= 180°

∴∠B and ∠G are a pair of supplementary angles.

v. m∠N + m∠J = 30° + 150°

= 180°

∴∠N and ∠J are a pair of supplementary angles.

Question 3.

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

Solution:

In ΔXYZ,

m∠X + m∠Y + m∠Z = 180° ….(Sum of the measure of the angles of a triangle is 180°)

∴m∠X + 90 + m∠Z = 180

∴m∠X + 90 + m∠Z – 90 = 180 – 90 ….(Subtracting 90 from both sides)

∴m∠X + m∠Z = 90°

∴∠X and ∠Z make a pair of 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.

Solution:

Let the measure of one angle be x°.

∴Measure of other angle = (x + 40)°

x + (x + 40) = 90 …(Since, the two angles are complementary)

∴ 2x + 40 – 40 = 90 – 40 ….(Subtracting 40 from both sides)

∴2x = 50

∴x = \(\frac { 50 }{ 2 }\)

∴x = 25

∴x + 40 = 25 + 40

= 65

∴The measures of the two angles is 25° and 65°.

Question 5.

₹PTNM is a rectangle. Write the names of the pairs of supplementary angles.

Solution:

Since, each angle of the rectangle is 90°.

∴ Pairs of supplementary angles are:

i. ∠P and ∠M

ii. ∠P and ∠N

iii. ∠P and ∠T

iv. ∠M and ∠N

v. ∠M and ∠T

vi. ∠N and ∠T

Question 6.

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

Solution:

Let the measure of the complement of ∠A be x° and the measure of its supplementary angle be y°.

m∠A + x = 90°

∴70 + x = 90

∴70 + x – 70 = 90 – 70 ….(Subtracting 70 from both sides)

∴x = 20

Since, x and y are supplementary angles.

∴x + y = 180

∴20 + y = 180

∴20 + y – 20 = 180 – 20 ….(Subtracting 20 from both sides)

∴y = 160

∴The measure of 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?

Solution:

Since, ∠A and ∠B are supplementary angles.

∴m∠A + m∠B = 180

∴m∠A + x + 20 = 180

∴m∠A + x + 20 – 20 = 180 – 20 ….(Subtracting 20 from both sides)

∴m∠A + x = 160

∴m∠A + x – x = 160 – x ….(Subtracting x from both sides)

∴m∠A = (160 – x)°

∴The measure of ∠A is (160 – x)°.

**Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 17 Intext Questions and Activities**

Question 1.

Observe the figure and answer the following questions. (Textbook pg. no. 26)

T is a point on line AB.

- What kind of angle is ∠ATB?
- What is its measure?

Solution:

- Straight angle
- 180°