**RD Sharma Solutions Class 9 Maths Chapter 8:** To have a comprehensive study and intelligent learning of lines and angles principles and techniques. RD Sharma Solutions Class 9 Maths Chapter 8 empowers you with a detailed study of exercises and solutions to prepare and practice the concepts around lines and angles effectively.

**Download RD Sharma Solutions Class 9 Maths Chapter 8 – Lines and Angles PDF**

RD Sharma Solutions Class 9 Maths Chapter 8

**Exercise-wise: RD Sharma Solutions Class 9 Maths Chapter 8 – Lines and Angles **

**Access answers of ****RD Sharma Solutions Class 9 Maths Chapter 8 **

### RD Sharma Solutions Class 9 Maths Lines and Angles Chapter 8 Exercise 8.1

**Question 1: Write the complement of each of the following angles:**

**(i)20 ^{0}**

**(ii)35 ^{0}**

**(iii)90 ^{0}**

**(iv) 77 ^{0}**

**(v)30 ^{0}**

**Solution**:

**(i)** The sum of an angle and its complement = 90^{0}

*Therefore, the complement of 20^{0 }= 90^{0} – 20^{0} = 70^{0}*

**(ii)** The sum of an angle and its complement = 90^{0}

*Therefore, the complement of 35° = 90° – 35° = 55*

**(iii)** The sum of an angle and its complement = 90^{0}

*Therefore, the complement of 90^{0} = 90^{0} – 90^{0} = 0^{0}*

**(iv)** The sum of an angle and its complement = 90^{0}

*Therefore, the complement of 77^{0} = 90° – 77^{0} = 13^{0}*

**(v)** The sum of an angle and its complement = 90^{0}

*Therefore, the complement of 30^{0} = 90^{0} – 30^{0} = 60^{0}*

**Question 2: Write the supplement of each of the following angles:**

**(i) 54 ^{0}**

**(ii) 132 ^{0}**

**(iii) 138 ^{0}**

**Solution:**

**(i)** The sum of an angle and its supplement = 180** ^{0}**.

*Therefore supplement of angle 54^{0} = 180^{0} – 54^{0} = 126^{0}*

**(ii)** The sum of an angle and its supplement = 180** ^{0}**.

*Therefore supplement of angle 132^{0} = 180^{0} – 132^{0} = 48^{0}*

**(iii)** The sum of an angle and its supplement = 180** ^{0}**.

*Therefore supplement of angle 138^{0 }= 180^{0 }– 138^{0} = 42^{0}*

**Question 3: If an angle is 28 ^{0} less than its complement, find its measure.**

**Solution:**

Let the measure of any angle is ‘ a ‘ degrees

Thus, its complement will be (90 – a)^{ 0}

So, the required angle = Complement of a – 28

a = ( 90 – a ) – 28

2a = 62

a = 31

*Hence, the angle measured is 31^{0}.*

**Question 4: If an angle is 30° more than one-half of its complement, find the measure of the angle.**

**Solution**:

Let an angle measured by ‘ a ‘ in degrees

Thus, its complement will be (90 – a)^{ 0}

Required Angle = 30** ^{0}** + complement/2

a = 30** ^{0}** + ( 90 – a )

**/ 2**

^{ 0}a + a/2 = 30** ^{0}** + 45

^{0}3a/2 = 75^{0}

a = 50^{0}

*Therefore, the measure of the required angle is 50^{0}.*

**Question 5: Two supplementary angles are in the ratio 4:5. Find the angles.**

**Solution**:

Two supplementary angles are in the ratio 4:5.

Let us say, the angles are 4a and 5a (in degrees)

Since angles are supplementary angles;

Which implies, 4a + 5a = 180^{0}

9a = 180^{0}

a = 20^{0}

Therefore, 4a = 4 (20) = 80** ^{0 }**and

5(a) = 5 (20) = 100^{0}

*Hence, the required angles are 80° and 100^{0}.*

**Question 6: Two supplementary angles differ by 48 ^{0}. Find the angles?**

**Solution**: Given: Two supplementary angles differ by 48** ^{0}**.

Consider a** ^{0}** to be one angle then its supplementary angle will be equal to (180 – a)

^{ 0}According to the question;

(180 – a ) – x = 48

(180 – 48 ) = 2a

132 = 2a

132/2 = a

Or a = 66^{0}

Therefore, 180 – a = 114^{0}

*Hence, the two angles are 66^{0} and 114^{0}.*

**Question 7: An angle is equal to 8 times its complement. Determine its measure.**

**Solution:** Given: Required angle = 8 times of its complement

Consider a** ^{0}** to be one angle then its complementary angle will be equal to (90 – a)

^{ 0}According to the question;

a = 8 times its complement

a = 8 ( 90 – a )

a = 720 – 8a

a + 8a = 720

9a = 720

a = 80

*Therefore, the required angle is 80^{0}.*

All luck to the students of RD Sharma Solutions CBSE Class 9. RD Sharma Solutions Chapter 8 is the perfect companion for Math Class 9 students to come out with flying Colours in their Mathematics exams. For any other doubts regarding the Class 9 Maths exams, ask in the comments.

**FAQs on RD Sharma Solutions Class 9 Maths Chapter 8 **

### From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 8?

You can find the download link from the above blog.

### How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 8?

You can download it for free.

### Can I access the RD Sharma Solutions for Class 9 Maths Chapter 8 PDF offline?

Once you have downloaded the PDF online, you can access it offline as well.