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# RD Sharma Solution for Class 9 Mathematics Chapter 10 - Lines and Angles

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 10 - Lines and Angles.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## RD Sharma Solution for Class 9 Mathematics Chapter 10 - Lines and Angles Page/Excercise 10.1

Solution 1

Solution 2

(i) 54°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 54° = 126°.

(ii) 132°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 132° = 48°.

(iii) 138°

Since, the sum of an angle and its supplement is 180°

Its supplement will be 180° - 138° = 42°.

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Let the measure of the angle be xo.

Its complement will be (90o - xo) and its supplement will be (180o - xo).

Supplement of thrice of the angle = (180o - 3xo)

According to the given information:

(90o - xo) = (180o - 3xo)

3x - x = 180 - 90

2x = 90

x = 45

Thus, the measure of the angle is 45o.

The measure of the angle is 45o

Solution 14

## RD Sharma Solution for Class 9 Mathematics Chapter 10 - Lines and Angles Page/Excercise 10.2

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Given that OR ⊥ PQ      POR = 90o          ⇒ POS  + SOR = 90o   ROS = 90 - POS                ... (1)       QOR = 90o                     (As OR ⊥ PQ)       QOS - ROS = 90o       ROS = QOS - 90o             ... (2)       On adding equations (1) and (2), we have       2 ROS = QOS - POS

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(i) True

(ii) False

(iii) False

(iv) True

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(i) obtuse.

(ii) 180o

(iii) uncommon

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## RD Sharma Solution for Class 9 Mathematics Chapter 10 - Lines and Angles Page/Excercise 10.4

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Let AB and CD be perpendicuar to line MN.

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Let AB and CD be perpendicuar to line MN.

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Consider the angles AOB and ACB.

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## RD Sharma Solution for Class 9 Mathematics Chapter 10 - Lines and Angles Page/Excercise 10.56

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Correct option (c)

Let one angle be θ

Then, its complementary = 90 - θ

According to question,

2θ = 3(90 - θ)

5θ = 270

θ = 54°

Then,  90 - θ° = 36°

Hence, the smaller angle is 36°.

Hence, correct option is (c).

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## Browse Study Material

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 10 - Lines and Angles for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 10 - Lines and Angles.

# Text Book Solutions

CBSE IX - Mathematics

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