NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2 is provided in this article. Class 10 Maths Chapter 8 Introduction to Trigonometry is included under Unit 5 Trigonometry of class 10 maths syllabus. Chapter 8 exercise 8.2 covers important questions based on trigonometric ratios of some specific angles.

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Check out NCERT solutions of other exercises of class 10 maths chapter 8 Introduction to Trigonometry:

Class 10 Chapter 8 Introduction to Trigonometry Related Links:

Trigonometry Table	Trigonometric Functions	Cosec Cot Formula
Some Applications of Trigonometry	Introduction to Trigonometry Important Formula	Heights and Distances
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CBSE Class 10 Maths Study Guides:

Math Formula	Math Study Notes	Trigonometry
Math MCQs	NCERT Solutions for Class 10 Maths	Difference between in Maths
Arithmetic	Mensuration	Calculus
Class 10 Notes	Number System	Geometry

CBSE X Related Questions

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
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2.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
4.
Find the H.C.F. and L.C.M. of 1530 and 2040.
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5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution
6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

CBSE X Related Questions

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
Find the H.C.F. and L.C.M. of 1530 and 2040.

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

CBSE X Related Questions

1.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.Find the H.C.F. and L.C.M. of 1530 and 2040.

5.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
Find the H.C.F. and L.C.M. of 1530 and 2040.

5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.