Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 5 Lines and Angles Ex 5.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1

1. Name each of the following as acute, obtuse, right straight or a reflex angle.

Question (i).

Answer:

Right angle

Question (ii).

Answer:

Obtuse angle

Question (iii).

Answer:

Straight angle

Question (iv).

Answer:

Reflex angle

Question (v).

Answer:

Obtuse angle

Question (vi).

Answer:

Acute angle.

2. Write the complement of each of the following angles :

Question (i).

53°

Answer:

Complement of 53°

= (90° – 53°) = 37°.

Question (ii).

90°

Answer:

Complement of 90°

= (90° – 90°) = 0°.

Question (iii).

85°

Answer:

Complement of 85°

= (90° – 85°) = 5°.

Question (iv).

\(\frac {4}{9}\) of a right angle

Answer:

Complement of \(\frac {4}{9}\) of a right angle

i. e. 40° = (90° – 40°) = 50°

Question (v).

0°

Answer:

Complement of 0° = (90° – 0°)

= 90°.

3. Write the supplement of each of the following angle :

Question (i).

55°

Answer:

Supplement of 55°

= (180° – 55°) = 125°.

Question (ii).

105°

Answer:

Supplement of 105°

= (180° – 105°) = 75°.

Question (iii).

100°

Answer:

Supplement of 100°

= (180° – 100°) = 80°.

Question (iv).

\(\frac {2}{3}\) of a right angle

Answer:

\(\frac {2}{3}\) of a right angle

= \(\frac {2}{3}\) × 90° = 60°.

∴ Supplement of 60°

= (180° – 60°) = 120°.

Question (v).

\(\frac {1}{3}\) of 270°.

Answer:

Supplement of \(\frac {1}{3}\) of 270° i.e. 90°

= (180°- 90°) = 90°.

4. Identify the following pairs of angles as complementary or supplementary.

Question (i).

65° and 115°

Answer:

Since 65° + 115° = 180°.

∴ It is a pair of supplementary angles.

Question (ii).

112° and 68°

Answer:

Since 112° + 68° = 180°

∴ It is a pair of supplementary angles.

Question (iii).

63° and 27°

Answer:

Since 63° + 27° = 90°

∴ It is a pair of complementary angles.

Question (iv).

45° and 45°

Answer:

Since 45° + 45° = 90°

∴ It is a pair of complementary angles.

Question (v).

130° and 50°

Answer:

Since 130° + 50° = 180°.

∴ It is a pair of supplementary angles.

5. Two complementary angles are in the ratio of 4 : 5, find the angles.

Solution:

Ratio of angles = 4 : 5

Let two complementary angles are 4x and 5x

Their sum = 90°

∴ 4x + 5x = 90°

9x = 90°

x = \(\frac{90^{\circ}}{9}\) = 10°

∴ 1st angle = 4x = 4 × 10° = 40°

2nd angle = 5x = 4 × 10° = 50°

6. Two supplementary angles are in the ratio of 5 : 13, find the angles.

Solution:

Ratio of two supplementary angles = 5 : 13

Let 5x and 13x are two supplementary angles

Since their sum = 180°

∴ 5x + 13x = 180°

18x = 180°

x = \(\frac{180^{\circ}}{18}\) = 10°

∴ 1st angle = 5x = 5 × 10° = 50°.

2nd angle = 13x = 13 × 10° = 130°

7. Find the angle which is equal to its complement.

Solution:

Let the angle be = x

Therefore its complement = 90° – x

Since the angle is equal to its complement

∴ x = 90° – x

or x + x = 90°

or 2x = 90°

or x = \(\frac{90^{\circ}}{2}\)

or x = 45°

Therefore the required angle is 45°.

8. Find the angle which is equal to its supplement.

Solution:

Let the angle be x

Therefore its supplement = 180° – x

Since the angle is equal to its supplement

∴ x = 180° – x

or x + x = 180°

or 2x = 180°

or x = \(\frac{180^{\circ}}{2}\) = 90°

Therefore, the required angle is 90°.

9. In the given figure, AOB is straight line. Find the measure of ∠AOC.

Solution:

In the given figure AOB is straight line (see Fig.)

∴ ∠AOB = 180°

∴ ∠AOC + ∠BOC = 180°

or ∠AOC + 50° = 180°

[∵ ∠BOC = 50° (given)]

∴∠AOC = 180° – 50°

= 130°

10. In the given figure, MON is straight line find.

(i) ∠MOP

(ii) ∠NOP

Solution:

Since MON is straight line (see Fig.)

∴ ∠MON = 180°

∴ ∠MOP + ∠NOP = 180°

[∵ ∠MOP = x + 20°

∠NOP = x + 40°]

or 2x + 60° = 180°

or 2x = 180° – 60°

or 2x = 120°

or x = \(\frac{120^{\circ}}{2}\) = 60°.

(i) ∠MOP = x + 20°

= 60° + 20°

= 80°

(ii) ∠NOP = x + 40°

= 60° + 40°

= 100°

11. Find the value of x, y and z in each of following.

Question (i).

Solution:

In fig (i)

x = 100°

(Vertically opposite angles)

y = 80°

(Vertically opposite angles)

Question (ii).

Solution:

In fig (ii)

z = 60°

(Vertically opposite angles)

∠y + 60° = 180° (Linear pair)

or ∠y = 180° – 60°

or ∠y = 120°

x = y

(Vertically opposite angles)

= 120°

12. Find the value of x, y, z and p in each of following.

Question (i).

Solution:

In fig (i)

45° + x + 35°= 180° (Linear pair)

or x + 80° = 180°

or x = 180° – 80°

or x = 100°

y = 45° (Vertically opposite angles)

Also 45° + z = 180° (Linear pair)

z = 180° – 45°

= 135°.

Hence x = 100°,

y = 45°,

z = 135°

Question (ii).

Solution:

In fig (ii)

p + 65° + 55° = 180° (Linear pair)

p + 120°= 180°

∴ p = 180° -120°

i. e. p = 60°

x = 55°

(Vertically opposite angles)

y = 65°

(Vertically opposite angles)

z = p

= 60°

(Vertically opposite angles)

Hence x = 55°,

y = 65°,

z = 60°,

p = 60°

13. Multiple Choice Questions :

Question (i).

If two angles are complementary then the sum of their measure is …………..

(a) 180°

(b) 90°

(c) 360°

(d) None of these.

Answer:

(b) 90°

Question (ii).

Two angles are called ………….. if the sum of their measures is 180°.

(a) supplementary

(b) complementary

(c) right

(d) none of these.

Answer:

(a) supplementary

Question (iii).

If two adjacent angles are supplementary then, they form a …………..

(a) right angle

(b) vertically opposite angles

(c) linear pair

(d) corresponding angles.

Answer:

(c) linear pair

Question (iv).

If two lines intersect at a point, the vertically opposite angles are always …………..

(a) equal

(b) zero

(c) 90°

(d) none of these.

Answer:

(a) equal