NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3 Solutions are based on the Angle Sum Property of a Triangle and the theorems relevant to it.

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Check out NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 6 Lines and Angles

Also check other Exercise Solutions of Class 9 Maths Chapter 6 Lines and Angles

Also Check:

Important Chapter Related Topics
Pairs of Angles	Parallel Lines and a Transversal	Collinear Points
Vertex	Linear Pair Axiom	Transversal
Corresponding Angles Axiom	Alternate Angles	Properties of Parallel Lines
Angle Formula	Obtuse Angle	Class 9 Maths Chapter 6 Lines and Angles MCQs
Adjacent Angles Vertical	Linear Pair of Angles	Intersecting Lines and Non Intersecting Lines

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9 \csc \theta + 1}{9 \csc \theta - 1}$ is
- $0$
- $\frac{80}{81}$
- $1$
- $\frac{82}{80}$
View Solution
2.
Arc $PQ$ subtends an angle $\theta$ at the centre of the circle with radius $6.3 \text{ cm}$. If $\text{Arc } PQ = 11 \text{ cm}$, then the value of $\theta$ is
- $10^{\circ}$
- $60^{\circ}$
- $45^{\circ}$
- $100^{\circ}$
View Solution
3.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is
- 34
- 43
- 38.5
- 41.5
View Solution
4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is
- $6$
- $4$
- $\pm 4$
- $3$
View Solution
5.
Assertion (A) : H.C.F. $(36 m^{2}, 18 m) = 18 m$, where $m$ is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
6.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?
- 144
- 2
- 420
- 272
View Solution

View All

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You can also check

NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 6 Lines and Angles

CBSE X Related Questions

1.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

2.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

3.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

6.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 6 Lines and Angles

CBSE X Related Questions

1.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

2.Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

3.Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

4.In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

6.There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

2.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

3.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

4.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

5.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

6.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?