**NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry:**Trigonometry is a branch of mathematics that deals with the study of triangles and their properties. It involves the study of the relationships between the sides and angles of triangles, and it has many applications in fields such as engineering, physics, architecture, and astronomy.

Trigonometry is based on the three fundamental trigonometric functions: sine (sin), cosine (cos), and tangent (tan). These functions relate the angles of a triangle to the ratios of the sides of the triangle. The sine function relates the opposite side of an angle to the hypotenuse, the cosine function relates the adjacent side to the hypotenuse, and the tangent function relates the opposite side to the adjacent side.

Trigonometry also includes the study of inverse trigonometric functions, which allow us to find the angles of a triangle given the ratios of its sides. Other important concepts in trigonometry include the Pythagorean theorem, which relates the sides of a right triangle, and the Law of Sines and Law of Cosines, which relate the sides and angles of any triangle.

Trigonometry has many practical applications, such as in the calculation of distances and angles in navigation, in the design of buildings and bridges, and in the analysis of sound and light waves. It is also essential in fields such as astronomy, where it is used to measure the distances and sizes of stars and planets.

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Post Contents

Introduction to Trigonometry Class 10 has total four exercises having 27 problems. Questions based on trigonometric ratios of specific angles, trigonometric identities and trigonometric ratios of complementary angles are the main topics that you will learn in this chapter. Trigonometry formulas play an important role in scoring 100% marks in board exams.

**Topics and Sub Topics in Class 10 Maths Chapter 8 Introduction to Trigonometry:**

Section Name |
Topic Name |

8 | Introduction to Trigonometry |

8.1 | Introduction |

8.2 | Trigonometric Ratios |

8.3 | Trigonometric Ratios Of Some Specific Angles |

8.4 | Trigonometric Ratios Of Complementary Angles |

8.5 | Trigonometric Identities |

8.6 | Summary |

## NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

The NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry Introduction to Ex 8.1 are part of the NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Introduction to Trigonometry |

Exercise |
Ex 8.1 |

Number of Questions Solved |
11 |

Category |
NCERT Solutions |

### What is Trigonometry

Trigonometry is a branch of mathematics that deals with the study of triangles and their properties. It involves the study of the relationships between the sides and angles of triangles, and it has many applications in fields such as engineering, physics, architecture, and astronomy.

Trigonometry is based on the three fundamental trigonometric functions: sine (sin), cosine (cos), and tangent (tan). These functions relate the angles of a triangle to the ratios of the sides of the triangle. The sine function relates the opposite side of an angle to the hypotenuse, the cosine function relates the adjacent side to the hypotenuse, and the tangent function relates the opposite side to the adjacent side.

Trigonometry also includes the study of inverse trigonometric functions, which allow us to find the angles of a triangle given the ratios of its sides. Other important concepts in trigonometry include the Pythagorean theorem, which relates the sides of a right triangle, and the Law of Sines and Law of Cosines, which relate the sides and angles of any triangle.

Trigonometry has many practical applications, such as in the calculation of distances and angles in navigation, in the design of buildings and bridges, and in the analysis of sound and light waves. It is also essential in fields such as astronomy, where it is used to measure the distances and sizes of stars and planets.

### Basic rules of Trigonometry

There are several basic rules of trigonometry that are important to understand. Some of the most important rules include:

- The Pythagorean theorem: In a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. This theorem can be written as a^2 + b^2 = c^2, where c is the hypotenuse, and a and b are the other two sides.
- The sine, cosine, and tangent ratios: These ratios relate the sides of a right triangle to its angles. The sine of an angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse, the cosine of an angle is equal to the length of the side adjacent to the angle divided by the length of the hypotenuse, and the tangent of an angle is equal to the length of the side opposite the angle divided by the length of the side adjacent to the angle.
- The reciprocal trigonometric functions: The reciprocal functions of sine, cosine, and tangent are cosecant, secant, and cotangent, respectively. These functions are defined as the reciprocals of their corresponding trigonometric functions.
- The Law of Sines: This law relates the ratios of the sides and angles of any triangle. Specifically, it states that the ratio of the length of a side of a triangle to the sine of the opposite angle is the same for all sides and angles of the triangle.
- The Law of Cosines: This law also relates the sides and angles of any triangle, but it is used to find the length of a side when the lengths of the other two sides and the included angle are known. The law states that the square of the length of a side of a triangle is equal to the sum of the squares of the other two sides minus twice their product times the cosine of the included angle.

These rules are fundamental to the study of trigonometry and are used in many practical applications, such as navigation, engineering, and physics.

### Use of Trigonometry in daily life

Trigonometry has numerous applications in daily life. Some examples include:

- Navigation: Trigonometry is used in navigation to determine the distance between two points, the direction of travel, and the position of objects. For example, pilots and sailors use trigonometry to navigate through the air or on the sea.
- Architecture and construction: Trigonometry is used in the design and construction of buildings, bridges, and other structures. Architects and engineers use trigonometry to calculate angles, heights, and distances to ensure that their designs are safe and functional.
- Surveying: Surveyors use trigonometry to measure land and create maps. They use angles and distances to calculate the size and shape of plots of land.
- Physics: Trigonometry is used in physics to study the movement of objects and waves. For example, physicists use trigonometry to analyze the motion of waves in water and sound waves in air.
- Astronomy: Trigonometry is used in astronomy to measure the distance and size of stars and planets. Astronomers use trigonometry to calculate the angle between the Earth and a star or planet and use that information to determine its distance.
- Sports: Trigonometry is used in sports such as baseball, soccer, and football to calculate the trajectory and distance of a ball when it is thrown or kicked.

Ttrigonometry plays a vital role in our daily lives and has many practical applications in various fields.

## Introduction to Trigonometry Ex 8.1 Questions And Solutions

### Ex 8.1 Class 10 Maths Question 1.

In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Solution:

Given that ∆ABC is a right angled triangle at B, with AB = 24 cm and BC = 7 cm.

(i) To find sin A and cos A, we need to first find the length of AC using the Pythagorean theorem.

AC² = AB² + BC² AC² = (24)² + (7)² AC² = 576 + 49 AC² = 625 AC = 25 cm

Now, we can use the definitions of sine and cosine to find their values:

sin A = opposite / hypotenuse = BC / AC = 7/25

cos A = adjacent / hypotenuse = AB / AC = 24/25

Therefore, sin A = 0.28 and cos A = 0.96.

(ii) To find sin C and cos C, we can use the fact that the sum of the angles in a triangle is 180 degrees, and we already know that angle B is 90 degrees.

So, angle C = 180 – 90 – angle A = 90 – angle A.

We can use the sine and cosine of angle A that we already found to find the sine and cosine of angle C:

sin C = sin(90 – A) = cos A = 0.96

cos C = cos(90 – A) = sin A = 0.28

Therefore, sin C = 0.96 and cos C = 0.28.

### Ex 8.1 Class 10 Maths Question 2.

In given figure, find tan P – cot R.

Solution:

You can also download the free **PDF of Chapter 8 Ex 8.1 Introduction to Trigonometry NCERT Solutions** or save the solution images and take the print out to keep it handy for your exam preparation.

### Ex 8.1 Class 10 Maths Question 3.

If sin A = 34 , calculate cos A and tan A.

Solution:

The value of sin A cannot be greater than 1, so it seems there may be an error in the given value of sin A = 34. Therefore, we cannot proceed with the calculation of cos A and tan A.

Please check the value of sin A and provide the correct value so that we can assist you further.

### Ex 8.1 Class 10 Maths Question 4.

Given 15 cot A = 8, find sin A and sec A.

Solution:

We can use the fact that cot A = 1 / tan A to rewrite the given equation:

15 cot A = 8 15 (1 / tan A) = 8 tan A = 15 / 8

Now, we can use the definitions of sine and cosine to find their values:

sin A = opposite / hypotenuse = 15 / (sqrt(15² + 8²)) = 0.78

cos A = adjacent / hypotenuse = 8 / (sqrt(15² + 8²)) = 0.62

Alternatively, we can use the fact that sec A = 1 / cos A to find sec A:

sec A = 1 / cos A = 1 / 0.62 = 1.61

Therefore, sin A is approximately 0.78 and sec A is approximately 1.61.

### Ex 8.1 Class 10 Maths Question 5.

Given sec θ = 1312 , calculate all other trigonometric ratios.

Solution:

We can use the definitions of trigonometric ratios and the fact that sec θ = 1 / cos θ to find the values of other trigonometric ratios:

sec θ = 1 / cos θ = 1312

cos θ = 1 / sec θ = 1 / 1312

sin θ = sqrt(1 – cos² θ) = sqrt(1 – (1/1312)²) = sqrt(1 – 1/1719424) = sqrt(1719423/1719424) ≈ 0.999999424

tan θ = sin θ / cos θ = (sqrt(1719423/1719424)) / (1/1312) = 1312 sqrt(1719423/1719424) ≈ 0.999999616

csc θ = 1 / sin θ = 1 / (sqrt(1719423/1719424)) ≈ 1.000000576

cot θ = 1 / tan θ = (1/1312) / (sqrt(1719423/1719424)) ≈ 1.000000384

Therefore, cos θ is approximately 0.000762, sin θ is approximately 0.999999424, tan θ is approximately 0.999999616, csc θ is approximately 1.000000576, and cot θ is approximately 1.000000384.

### Ex 8.1 Class 10 Maths Question 6.

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:

We can use the fact that the cosine function is a decreasing function in the interval [0, 180 degrees] for acute angles to prove that if cos A = cos B, then ∠A = ∠B.

Since 0 degrees < ∠A, ∠B < 90 degrees, we know that cos A and cos B are both positive. Also, since cos A = cos B, we have:

cos A – cos B = 0

Using the identity for the difference of cosines, we can rewrite this as:

-2 sin[(A+B)/2] sin[(A-B)/2] = 0

Since sin[(A+B)/2] and sin[(A-B)/2] are both positive for acute angles A and B, we can conclude that sin[(A-B)/2] must be zero. This implies that either A = B or A + B = 180 degrees.

However, since A and B are both acute angles, we have A + B < 180 degrees, so the only possibility is that A = B. Therefore, we have shown that if cos A = cos B, then ∠A = ∠B for acute angles A and B

### Ex 8.1 Class 10 Maths Question 7.

If cot θ = 78, evaluate:

(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)

(ii) cot²θ

Solution:

We can use the identity 1 + cot² θ = csc² θ to find the value of cosecant and then use the given trigonometric identities to solve the given expressions:

cot θ = 1/tan θ, so tan θ = 1/cot θ = 1/78

Using the Pythagorean identity for tangent and the fact that θ is acute, we get:

tan θ = sin θ / cos θ = 1 / 78

sin θ = tan θ cos θ = (1/78) (sqrt(1 + tan² θ)) = (1/78) (sqrt(1 + (1/78²)))

cos θ = sqrt(1 – sin² θ) = sqrt(1 – (1/78²))

(i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)

Substituting the values of sin θ and cos θ, we get:

(1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ) = [(1 + (1/78²)) (1 – (1/78²))] [(1 + sqrt(1 – (1/78²))) (1 – sqrt(1 – (1/78²)))]

Simplifying this expression, we get:

(1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ) = [(78² + 1)/(78²)] [(78 + sqrt(6083))/(78 – sqrt(6083))] [(78 – sqrt(6083))/(78 + sqrt(6083))]

(1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ) = (6083/6084) = 7/8 (approx)

(ii) cot²θ

We can use the identity cot² θ + 1 = csc² θ to find the value of cosecant and then use the given value of cot θ to solve the expression:

cot² θ + 1 = csc² θ

cot² θ = csc² θ – 1 = (1/sin² θ) – 1

Substituting the value of sin θ, we get:

cot² θ = (1/(1/78²)) – 1 = 78² – 1 = 6083

Therefore, cot² θ = 6083.

### Ex 8.1 Class 10 Maths Question 8.

If 3 cot A = 4, check whether 1−tan2A1+tan2A = cos² A – sin² A or not.

Solution:

We can start by finding the value of tangent and cotangent of angle A using the given information:

3 cot A = 4

cot A = 4/3

tan A = 1/cot A = 3/4

Now, we can substitute these values in the expression:

(1 – tan² A)/(1 + tan² A) = cos² A – sin² A

[(1 – (3/4)²)/(1 + (3/4)²)] = cos² A – sin² A

[(1 – 9/16)/(1 + 9/16)] = cos² A – sin² A

[(7/16)/(25/16)] = cos² A – sin² A

7/25 = cos² A – sin² A

Now, we can use the identities cos² A + sin² A = 1 and cos² A – sin² A = cos(2A) to simplify the expression further:

cos² A – sin² A = cos(2A)

7/25 = cos(2A)

We can then use the inverse cosine function to find the value of 2A:

2A = cos⁻¹(7/25)

A = 1/2 cos⁻¹(7/25)

Therefore, the expression 1 – tan² A / (1 + tan² A) does not simplify to cos² A – sin² A for the given value of 3 cot A = 4.

### Ex 8.1 Class 10 Maths Question 9.

In triangle ABC, right angled at B, if tan A = 1√3, find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

Since triangle ABC is a right-angled triangle with angle A, we have:

tan A = opposite side / adjacent side = BC / AB

Given tan A = 1/√3, we can solve for the sides of the triangle:

BC / AB = 1/√3

BC = AB / √3

Using the Pythagorean theorem, we get:

AC² = AB² + BC²

AC² = AB² + (AB² / 3)

AC² = (4/3) AB²

AC = AB √(4/3)

AC = AB (2/√3)

(i) sin A cos C + cos A sin C

We know that sin A = opposite / hypotenuse and cos C = adjacent / hypotenuse. Therefore:

sin A cos C + cos A sin C = (opposite / hypotenuse) * (adjacent / hypotenuse) + (adjacent / hypotenuse) * (opposite / hypotenuse)

sin A cos C + cos A sin C = (opposite * adjacent) / hypotenuse² + (adjacent * opposite) / hypotenuse²

sin A cos C + cos A sin C = 2 (opposite * adjacent) / hypotenuse²

Substituting the values of opposite and adjacent, we get:

sin A cos C + cos A sin C = 2 (AB/√3) * (AB/2) / (AB² * 4/3)

sin A cos C + cos A sin C = √3 / 4

(ii) cos A cos C – sin A sin C

We know that cos A = adjacent / hypotenuse and sin C = opposite / hypotenuse. Therefore:

cos A cos C – sin A sin C = (adjacent / hypotenuse) * (adjacent / hypotenuse) – (opposite / hypotenuse) * (opposite / hypotenuse)

cos A cos C – sin A sin C = (adjacent² – opposite²) / hypotenuse²

Substituting the values of adjacent and opposite, we get:

cos A cos C – sin A sin C = [(AB/2)² – (AB/√3)²] / (AB²)

cos A cos C – sin A sin C = (AB² / 12) / (AB²)

cos A cos C – sin A sin C = 1/12

Therefore, sin A cos C + cos A sin C = √3 / 4 and cos A cos C – sin A sin C = 1/12.

### Ex 8.1 Class 10 Maths Question 10.

In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution:

Using Pythagoras theorem in right triangle PQR, we have:

PR² = PQ² + QR²

PR² = 5² + QR²

PR² = 25 + QR²

PR = sqrt(25 + QR²)

Also, we are given that PR + QR = 25, so we can substitute PR to get:

sqrt(25 + QR²) + QR = 25

Simplifying this equation, we get:

QR² – 10 QR + 25 = 0

(QR – 5)² = 0

QR = 5

Now, using the trigonometric ratios in right triangle PQR, we have:

sin P = opposite / hypotenuse = QR / PR = 5 / sqrt(25 + 5²) = 5 / 10 = 1/2

cos P = adjacent / hypotenuse = PR / PQ = sqrt(25 + 5²) / 5 = 2sqrt(6)

tan P = opposite / adjacent = QR / PQ = 5 / 5 = 1

Therefore, sin P = 1/2, cos P = 2sqrt(6) and tan P = 1

### Ex 8.1 Class 10 Maths Question 11.

State whether the following statements are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 125 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 43 for some angle.

Solution:

(i) False. The value of tan A can be greater than or equal to 1, depending on the value of angle A. For example, tan 45° = 1.

(ii) False. The value of sec A can never be greater than or equal to 1, since secant is the reciprocal of cosine and cosine takes values between -1 and 1. Therefore, the maximum value of secant is -1 and the minimum value is 1.

(iii) False. The abbreviation used for the cosecant of angle A is csc A. Cos A stands for the cosine of angle A.

(iv) False. Cot A is not the product of cot and A. It is a trigonometric function which represents the ratio of the adjacent side to the opposite side of an angle in a right triangle.

(v) False. The sine function takes values between -1 and 1, so the value of sin θ cannot be 43.

## Define Trigonometric

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles, particularly right triangles. It involves the study of the six trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent, as well as their inverses and reciprocal functions. Trigonometry is used extensively in fields such as engineering, physics, navigation, and architecture, among others, where it is applied to problems involving angles and distances. The study of trigonometry is important for understanding and solving problems related to triangles and their applications in the real world.

### Trigonometric Ratios

Trigonometric ratios are ratios of the sides of a right triangle with respect to its acute angles. The three primary trigonometric ratios are sine, cosine, and tangent, which are abbreviated as sin, cos, and tan, respectively. These ratios are defined as follows:

- Sine (sin) is the ratio of the length of the opposite side to the length of the hypotenuse. That is, sin A = opposite/hypotenuse.
- Cosine (cos) is the ratio of the length of the adjacent side to the length of the hypotenuse. That is, cos A = adjacent/hypotenuse.
- Tangent (tan) is the ratio of the length of the opposite side to the length of the adjacent side. That is, tan A = opposite/adjacent.

In addition to these three primary trigonometric ratios, there are also three reciprocal trigonometric ratios: cosecant (csc), secant (sec), and cotangent (cot), which are the inverse of the primary trigonometric ratios. That is, csc A = 1/sin A, sec A = 1/cos A, and cot A = 1/tan A.

These ratios are used to solve problems involving triangles and angles, and they are fundamental to the study of trigonometry.

### Trigonometric Ratios angle Table

Here is a table of the values of the six primary trigonometric ratios (sin, cos, tan, csc, sec, cot) for angles 0° to 90°:

Angle (deg) | sin | cos | tan | csc | sec | cot |
---|---|---|---|---|---|---|

0° | 0 | 1 | 0 | N/A | 1 | N/A |

15° | 0.2588 | 0.9659 | 0.2679 | 3.8637 | 1.0353 | 3.7321 |

30° | 0.5 | 0.866 | 0.5774 | 2 | 1.1547 | 1.7321 |

45° | 0.7071 | 0.7071 | 1 | 1.4142 | 1.4142 | 1 |

60° | 0.866 | 0.5 | 1.7321 | 1.1547 | 2 | 0.5774 |

75° | 0.9659 | 0.2588 | 3.7321 | 1.0366 | 3.8538 | 0.2679 |

90° | 1 | 0 | N/A | N/A | N/A | 0 |

Note that the values of the trigonometric ratios are dependent on the angle, and that the ratios are undefined for angles of 90° and 270° for tangent and cotangent, and for angles of 0° for cosecant and secant.

### Trigonometric Ratios for Complementary Angles

Let’s consider an acute angle A and its complement angle B. Then, we have:

- sin B = cos A
- cos B = sin A
- tan B = cot A
- cot B = tan A
- sec B = csc A
- csc B = sec A

These relationships follow from the definitions of the trigonometric ratios and the fact that the sum of the angles A and B is 90 degrees.

For example, if we know that the sine of angle A is 0.6, then we can find the cosine of its complement angle B as follows:

- sin B = cos A
- cos B = sin A (since sin B = sin (90 – A) = cos A)
- cos B = 0.6

Similarly, if we know that the tangent of angle A is 1.5, then we can find the cotangent of its complement angle B as follows:

- tan B = cot A
- cot B = tan A (since tan B = tan (90 – A) = cot A)
- cot B = 1/1.5 = 0.6667

These relationships can be useful in trigonometry problems involving complementary angles.

### Trigonometric Identities

- Pythagorean identity: sin^2θ + cos^2θ = 1 This identity relates the values of sine and cosine for any angle θ in a right triangle. It states that the square of the sine of an angle plus the square of the cosine of that angle is always equal to one.
- Reciprocal identities:

- cscθ = 1/sinθ
- secθ = 1/cosθ
- cotθ = 1/tanθ These identities relate the primary trigonometric ratios (sin, cos, and tan) to their reciprocals (csc, sec, and cot).

- Quotient identity: tanθ = sinθ/cosθ This identity expresses the tangent of an angle in terms of its sine and cosine.
- Even/odd identities:

- sin(-θ) = -sinθ
- cos(-θ) = cosθ
- tan(-θ) = -tanθ These identities relate the trigonometric functions for a negative angle to those for the corresponding positive angle.

- Double angle identities:

- sin2θ = 2sinθcosθ
- cos2θ = cos^2θ – sin^2θ
- tan2θ = 2tanθ/1-tan^2θ These identities express the values of the trigonometric functions for an angle that is twice the original angle in terms of the original angle.

- Sum and difference identities:

- sin(α ± β) = sinαcosβ ± cosαsinβ
- cos(α ± β) = cosαcosβ ∓ sinαsinβ
- tan(α ± β) = (tanα ± tanβ)/(1 ∓ tanαtanβ) These identities express the values of the trigonometric functions for the sum or difference of two angles in terms of the values of the functions for the individual angles.

These identities are just a few examples of the many trigonometric identities that exist.

### Tips and Tricks to solve Trigonometry

Here are some tips and tricks to help you solve trigonometry problems more efficiently:

- Memorize the basic trigonometric ratios: Knowing the ratios of sine, cosine, and tangent of common angles (such as 0, 30, 45, 60, and 90 degrees) can save you a lot of time when solving problems.
- Understand the properties of right triangles: Trigonometry is primarily concerned with the relationships between the sides and angles of right triangles. Understanding the properties of right triangles, such as the Pythagorean theorem, can help you solve problems more easily.
- Use the unit circle: The unit circle is a circle with a radius of 1 that is used to define the values of sine and cosine for all angles. Understanding the unit circle can help you quickly determine the values of sine and cosine for any angle.
- Use the appropriate trigonometric function: Each trigonometric function (sine, cosine, tangent) is used to relate different pairs of sides in a right triangle. Choosing the appropriate function for the given problem can simplify the calculations.
- Simplify expressions: Trigonometric expressions can often be simplified using algebraic manipulation or trigonometric identities. Simplifying expressions can make them easier to work with and can reveal patterns that can help you solve the problem.
- Practice, practice, practice: Like any skill, solving trigonometry problems requires practice. Work through as many problems as you can to become more comfortable with the concepts and techniques involved.

By following these tips and tricks, you can become more efficient and effective at solving trigonometry problems.

### Trigonometric examples

- Finding the length of a side in a right triangle: If we know the values of two sides of a right triangle and want to find the length of the third side, we can use trigonometry. For example, if we know that one angle of a right triangle is 30 degrees and that the side opposite that angle has length 5, we can use the sine function to find the length of the hypotenuse: sin(30) = opposite/hypotenuse sin(30) = 5/hypotenuse hypotenuse = 5/sin(30) ≈ 10.0
- Solving for an angle in a right triangle: If we know the lengths of two sides of a right triangle and want to find the value of one of the acute angles, we can use trigonometry. For example, if we know that one side of a right triangle has length 3 and the hypotenuse has length 5, we can use the inverse sine function to find the value of the angle opposite the side of length 3: sin(θ) = opposite/hypotenuse sin(θ) = 3/5 θ = sin^-1(3/5) ≈ 36.9 degrees
- Finding the height of an object using angle of elevation: If we know the angle of elevation from a point on the ground to the top of an object and the distance from the point to the object, we can use trigonometry to find the height of the object. For example, if we stand 50 feet from a building and look up at a 30 degree angle to see the top of the building, we can use the tangent function to find the height of the building: tan(30) = opposite/adjacent tan(30) = height/50 height = 50*tan(30) ≈ 28.9 feet

These are just a few examples of how trigonometry can be used to solve problems in real-world situations.

## FAQs About NCERT Solutions for Class 10 Maths Chapter 8 Trigonometry

Here are some frequently asked questions about trigonometry:

**What is the difference between sine, cosine, and tangent?**Sine, cosine, and tangent are the three primary trigonometric functions. Sine (sin) is the ratio of the opposite side to the hypotenuse in a right triangle. Cosine (cos) is the ratio of the adjacent side to the hypotenuse, and tangent (tan) is the ratio of the opposite side to the adjacent side.**What is the Pythagorean theorem, and how is it related to trigonometry?**The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This theorem is used in trigonometry to relate the values of the sine and cosine functions for any angle in a right triangle.**What are the trigonometric identities, and why are they important?**Trigonometric identities are equations that are true for all values of the variables involved. They are important in trigonometry because they allow us to simplify expressions and solve equations involving trigonometric functions.**What are radians, and how are they related to degrees?**Radians are a unit of measurement for angles that is based on the radius of a circle. One radian is equal to the angle that subtends an arc of length equal to the radius of the circle. Radians are related to degrees by the conversion factor of 180 degrees/pi radians.**How is trigonometry used in real-world applications?**Trigonometry is used in a wide range of fields, including engineering, physics, architecture, and navigation. It is used to solve problems involving angles and distances, such as finding the height of a building or the distance between two points.

**Here are some frequently asked questions about NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry:**

**Why is it important to study Trigonometry in Class 10?**Trigonometry is an important branch of mathematics that has numerous applications in real-life situations. It helps in solving problems related to heights and distances, angles of elevation and depression, and navigation. Studying Trigonometry in Class 10 lays a strong foundation for higher studies in mathematics and related fields.**What topics are covered in NCERT Solutions for Class 10 Maths Chapter 8?**NCERT Solutions for Class 10 Maths Chapter 8 cover topics such as Trigonometric ratios, Trigonometric ratios of complementary angles, Heights and Distances, and applications of Trigonometry.**Are NCERT Solutions for Class 10 Maths Chapter 8 sufficient for the Board exams?**NCERT Solutions for Class 10 Maths Chapter 8 are prepared in accordance with the latest CBSE guidelines and cover all the important topics. These solutions are well-structured and provide a step-by-step explanation for all the questions. Solving the NCERT Solutions for Class 10 Maths Chapter 8 will help students to score well in the Board exams.**How can I improve my Trigonometry skills?**To improve your Trigonometry skills, you should practice solving problems from textbooks and previous year question papers. You can also refer to online resources such as videos and quizzes to strengthen your understanding of the concepts. Additionally, attending Math classes regularly and seeking help from your teachers and peers can also be beneficial.**How can I download NCERT Solutions for Class 10 Maths Chapter 8?**NCERT Solutions for Class 10 Maths Chapter 8 can be downloaded from the official website of NCERT or other educational websites. Many mobile apps also provide NCERT Solutions that can be downloaded for free.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry PDF Download

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables