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

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). Right angle

Question (ii). Obtuse angle

Question (iii). Straight angle

Question (iv). Reflex angle Question (v). Obtuse angle

Question (vi). Acute angle.

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

Question (i).
53°
Complement of 53°
= (90° – 53°) = 37°.

Question (ii).
90°
Complement of 90°
= (90° – 90°) = 0°.

Question (iii).
85°
Complement of 85°
= (90° – 85°) = 5°. Question (iv).
$$\frac {4}{9}$$ of a right angle
Complement of $$\frac {4}{9}$$ of a right angle
i. e. 40° = (90° – 40°) = 50° Question (v).

Complement of 0° = (90° – 0°)
= 90°.

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

Question (i).
55°
Supplement of 55°
= (180° – 55°) = 125°.

Question (ii).
105°
Supplement of 105°
= (180° – 105°) = 75°.

Question (iii).
100°
Supplement of 100°
= (180° – 100°) = 80°. Question (iv).
$$\frac {2}{3}$$ of a right angle
$$\frac {2}{3}$$ of a right angle
= $$\frac {2}{3}$$ × 90° = 60°.
∴ Supplement of 60°
= (180° – 60°) = 120°.

Question (v).
$$\frac {1}{3}$$ of 270°.
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°
Since 65° + 115° = 180°.
∴ It is a pair of supplementary angles.

Question (ii).
112° and 68°
Since 112° + 68° = 180°
∴ It is a pair of supplementary angles.

Question (iii).
63° and 27°
Since 63° + 27° = 90°
∴ It is a pair of complementary angles. Question (iv).
45° and 45°
Since 45° + 45° = 90°
∴ It is a pair of complementary angles.

Question (v).
130° and 50°
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.
(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.
(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.