Students often donâ€™t clear their doubts in the class when solved by the teacher owing to shyness. This mainly occurs in a subject like Mathematics because in other subjects the concepts can be understood by re-reading the chapter. But, in Mathematics, the students need to understand the methods which are used in solving problems. In order to help students, we provide exercise wise answers to all the chapters in PDF format. By referring the PDF, students can understand the methods used in solving the problems. RS Aggarwal Solutions for Class 9 Maths Chapter 7 Lines and Angles Exercise 7A are provided here.

## RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7A Download PDF

## Access RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7A

__Exercise 7(A)__

**1. Define the following terms:**

**(i) Angle**

**(ii) Interior of an angle**

**(iii) Obtuse angle**

**(iv) Reflex angle**

**(v) Complementary angles**

**(vi) Supplementary angles**

**Solution:**

(i) Angle â€“ When two rays originate from the same end point, then an angle is formed.

(ii) Interior of an angle â€“ The interior of âˆ BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B.

(iii)Obtuse angle â€“ An angle whose measure is more than 90^{o} but less than 180^{o} is called an obtuse angle.

(iv) Reflex angle â€“ An angle whose measure is more than 180^{o} but less than 360^{o} is called a reflex angle.

(v) Complementary angles â€“ Two angles are said to be complementary, if the sum of their measure is 90^{o}.

(vi) Supplementary angles â€“ Two angles are said to be supplementary if the sum of their measures is 180^{o}.

**2. Find the complement of each of the following angles:**

**(i) 55 ^{o}**

**(ii) 16 ^{o}**

**(iii) 90 ^{o}**

**(iv) 2/3 of a right angle**

**Solution:**

(i) We know that the complement of 55^{o} can be written as

55^{o} = 90^{o} – 55^{o} = 35^{o}

(ii) We know that the complement of 16^{o} can be written as

16^{o} = 90^{o} – 16^{o} = 74^{o}

(iii) We know that the complement of 90^{o} can be written as

90^{o} = 90^{o} – 90^{o} = 0^{o}

(iv) We know that 2/3 of a right angle can be written as 2/3 Ã— 90^{o} = 60^{o}

60^{o }= 90^{o} – 60^{o} = 30^{o}

**3. Find the supplement of each of the following angles:**

**(i) 42 ^{o}**

**(ii) 90 ^{o}**

**(iii) 124 ^{o}**

**(iv) 3/5 of a right angle**

**Solution:**

(i) We know that the supplement of 42^{o} can be written as

42^{o} = 180^{o} – 42^{o} = 138^{o}

(ii) We know that the supplement of 90^{o} can be written as

90^{o} = 180^{o} – 90^{o} = 90^{o}

(iii) We know that the supplement of 124^{o} can be written as

124^{o} = 180^{o} – 124^{o} = 56^{o}

(iv) We know that 3/5 of a right angle can be written as 3/5 Ã— 90^{o} = 54^{o}

54^{o }= 180^{o} – 54^{o} = 126^{o}

**4. Find the measure of an angle which is**

**(i) Equal to its complement**

**(ii) Equal to its supplement**

**Solution:**

(i) Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

To find that the measure of an angle is equal to its complement

We get

x^{o} = 90^{o} – x^{o}

We can also write it as

x + x = 90

So we get

2x = 90

By division we get

x = 90/2

x^{o} = 45^{o}

Therefore, the measure of an angle which is equal to its complement is 45^{o}

(ii) Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

To find that the measure of an angle is equal to its complement

We get

x^{o} = 180^{o} – x^{o}

We can also write it as

x + x = 180

So we get

2x = 180

By division we get

x = 180/2

x^{o} = 90^{o}

Therefore, the measure of an angle which is equal to its complement is 90^{o}

**5. Find the measure of an angle which is 36 ^{o} more than its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

x^{o} = (90^{o} – x^{o}) + 36^{o}

We can also write it as

x + x = 90 + 36

So we get

2x = 126

By division we get

x = 126/2

x^{o} = 63^{o}

Therefore, the measure of an angle which is 36^{o} more than its complement is 63^{o}.

**6. Find the measure of an angle which is 30 ^{o} less than its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

x^{o} = (180^{o} – x^{o}) â€“ 30^{o}

We can also write it as

x + x = 180 – 30

So we get

2x = 150

By division we get

x = 150/2

x^{o} = 75^{o}

Therefore, the measure of an angle which is 30^{o} more than its supplement is 75^{o}.

**7. Find the angle which is four times its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o}

x^{o} = 4(90^{o} – x^{o})

We can also write it as

x = 360 â€“ 4x

So we get

5x = 360

By division we get

x = 360/5

x^{o} = 72^{o}

Therefore, the angle which is four times its complement is 72^{o}.

**8. Find the angle which is five times its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the supplement can be written as 180^{o} – x^{o}

x^{o} = 5(180^{o} – x^{o})

We can also write it as

x = 900 â€“ 5x

So we get

6x = 900

By division we get

x = 900/6

x^{o} = 150^{o}

Therefore, the angle which is five times its supplement is 150^{o}.

**9. Find the angle whose supplement is four times its complement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o} and the supplement can be written as 180^{o} – x^{o}

180^{o} – x^{o} = 4(90^{o} – x^{o})

We can also write it as

180^{o} – x^{o} = 360^{o} – 4x^{o}

So we get

4x^{o} – x^{o} = 360^{o} – 180^{o}

3x^{o} = 180^{o}

By division we get

x^{o} = 180/3

x^{o} = 60^{o}

Therefore, the angle whose supplement is four times its complement is 60^{o}.

**10. Find the angle whose complement is one third of its supplement.**

**Solution:**

Consider the required angle as x^{o}

We know that the complement can be written as 90^{o} – x^{o} and the supplement can be written as 180^{o} – x^{o}

90^{o} – x^{o} = 1/3(180^{o} – x^{o})

We can also write it as

90^{o} – x^{o} = 60^{o} â€“ (1/3) x^{o}

So we get

x^{o} â€“ (1/3)x^{o} = 90^{o} – 60^{o}

(2/3) x^{o} = 30^{o}

By division we get

x^{o} = ((30Ã—3)/2)

x^{o} = 45^{o}

Therefore, the angle whose complement is one third of its supplement is 45^{o}.

**11. Two complementary angles are in the ratio 4:5. Find the angles.**

**Solution:**

Consider the required angle as x^{o} and 90^{o} – x^{o}

According to the question it can be written as

x^{o}/ 90^{o} – x^{o} = 4/5

By cross multiplication we get

5x = 4 (90 â€“ x)

5x = 360 â€“ 4x

On further calculation we get

5x + 4x = 360

9x = 360

By division

x = 360/9

So we get

x = 40

Therefore, the angles are 40^{o} and 90^{o} – x^{o} = 90^{o} – 40^{o} = 50^{o}

**12. Find the value of x for which the angles (2x â€“ 5) ^{ o} and (x â€“ 10)^{ o} are the complementary angles.**

**Solution:**

It is given that (2x â€“ 5)^{ o} and (x â€“ 10)^{ o} are the complementary angles.

So we can write it as

(2x â€“ 5)^{ o} + (x â€“ 10)^{ o} = 90^{o}

2x â€“ 5^{o} + x â€“ 10^{o} = 90^{o}

On further calculation

3x – 15^{o} = 90^{o}

So we get

3x = 105^{o}

By division

x = 105/3

x = 35^{o}

Therefore, the value of x for which the angles (2x â€“ 5)^{ o} and (x â€“ 10)^{ o} are the complementary angles is 35^{o}.

### Access other exercise solutions of Class 9 Maths Chapter 7: Lines and Angles

Exercise 7B Solutions 16 Questions

Exercise 7C Solutions 24 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles Exercise 7A

RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Exercise 7A is the first exercise which contains examples and problems which are solved by experts based on complementary and supplementary angles.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles Exercise 7A

- The solutions are prepared by experts keeping in mind the understanding abilities among the students.
- Self analysis can be done by the students to understand the areas of higher importance based on weightage in the exam.
- RS Aggarwal solutions contain answers in descriptive manner which help students grasp the concepts faster.
- The problems are solved using the easiest methods and diagrammatic representation helps students to understand the methods.