NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon. Early astronomers used it to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometric concepts

The Introduction to Trigonometry Class 10th chapter introduces the basic trigonometric ratios-sine, cosine, and tangent-based on right-angled triangles. Through NCERT Class 10 Maths Solutions, students will easily find the step-by-step solution for all problems present in the NCERT textbook so that they can study how to solve those problems accurately. Class 10 Maths Chapter 8 Solutions are designed to strengthen students' conceptual understanding and problem-solving skills, ensuring better preparation for exams.

1.0NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry : Free PDF

Students can download NCERT Solutions pdf for class 10 Maths Chapter 8 from the link below: 

2.0NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry: All Exercises

Solving NCERT Textbook exercises can help students grasp basic understanding of the key concepts as it provides a wide range of problems covering all the topics. 

3.0What Will Students Learn in Chapter 8: Introduction to Geometry?

• In Maths Class 10 Introduction To Trigonometry chapter, you will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle.
• Students will also learn about the relationships between these ratios and how to apply them to solve various problems involving angles and sides of triangles
• Discussions will be restricted to acute angles only. However, these ratios can be extended to other angles also.
• You will also define the trigonometric ratios for angles of measure 0° and 90°.
• Calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

4.0Important Topics of Class 10 Math Chapter 8 Introduction of Trigonometry

Chapter 8, Introduction to Trigonometry, introduces students to the fundamental trigonometric ratios in a right-angled triangle: sine (sin), cosine (cos), and tangent (tan), as well as their reciprocals: cosecant (csc), secant (sec), and cotangent (cot). The chapter also covers trigonometric values for specific angles and important identities like ${\sin }^2\theta +{\cos }^2\theta =1$. These concepts provide a foundation for solving real-life problems involving angles and distances.

1. Trigonometric Ratios
2. Reciprocal Trigonometric Ratios
3. Trigonometric Identities
4. Quotient Relations
5. Power of Trigonometric Ratios
6. Trigonometric Values for Specific Angles

5.0General Outline for Class 10 Maths Chapter 8: Introduction to Trigonometry

Understanding Key Concepts

Trigonometric Ratios: These ratios relate the angles of a right-angled triangle to the lengths of its sides. For a right triangle with an angle θ:

1. Sine ($\sin \theta$) = $\frac{\text{Opposite}}{\text{Hypotenuse}}$
2. Cosine $(\cos \theta )=\frac{\text{Adjacent}}{\text{Hypotenuse}}$
3. Tangent $(\tan \theta )=\frac{\text{Opposite}}{\text{Adjacent}}$
4. Reciprocal ratios:

• Cosecant $(\csc \theta )=\frac{1}{\sin \theta }$
• Secant $(\sec \theta )=\frac{1}{\cos \theta }$
• Cotangent $(\cot \theta )=\frac{1}{\tan \theta }$

Trigonometric Ratios of Specific Angles

Standard Angles:

• $\sin 30^{\circ }=\frac{1}{2},\cos 30^{\circ }=\frac{\sqrt{3}}{2},\tan 30^{\circ }=\frac{1}{\sqrt{3}}$
• $\sin 45^{\circ }=\frac{\sqrt{2}}{2},\cos 45^{\circ }=\frac{\sqrt{2}}{2},\tan 45^{\circ }=1$
• $\sin 60^{\circ }=\frac{\sqrt{3}}{2},\cos 60^{\circ }=\frac{1}{2},\tan 60^{\circ }=\sqrt{3}$
• $\sin 90^{\circ }=1,\cos 90^{\circ }=0,\tan 90^{\circ }=\text{undefined}$

Important Theorems and Formulas

1. Pythagoras Theorem: ${\sin }^2\theta +{\cos }^2\theta =1$
2. Basic Trigonometric Identity:

• $\tan \theta =\frac{\sin \theta }{\cos \theta }$
• ${\sec }^2\theta =1+{\tan }^2\theta$
• ${\csc }^2\theta =1+{\cot }^2\theta$

3. Trigonometric Ratios in Different Quadrants

• In the first quadrant (0° to 90°), all trigonometric ratios are positive.
• In the second quadrant (90° to 180°), sine and cosecant are positive.
• In the third quadrant (180° to 270°), tangent and cotangent are positive.
• In the fourth quadrant (270° to 360°), cosine and secant are positive.

Steps for Solving Trigonometry Problems

Step 1: Identify the given information (angle, sides of the triangle, etc.).

Step 2: Label the sides of the triangle correctly relative to the given angle.

• Opposite side: The side opposite to the given angle.
• Adjacent side: The side next to the angle, excluding the hypotenuse.
• Hypotenuse: The longest side opposite the right angle.

Step 3: Choose the appropriate trigonometric ratio (sine, cosine, tangent, etc.) based on the sides involved in the problem.

Step 4: Apply the known values and solve the equation.

Step 5: If the problem asks for a specific angle, use the inverse trigonometric functions$({\sin }^{-1},{\cos }^{-1},{\tan }^{-1})$ to find the angle.

Step 6: Verify the solution (check if the side lengths satisfy the triangle’s properties or trigonometric identities).

Common Problem Types

• Finding Unknown Sides: Given one side and an angle, use the appropriate trigonometric ratio to find unknown sides of a right-angled triangle.
• Finding Angles: Given two sides, use the inverse trigonometric functions to find the angles of a triangle.
• Using Trigonometric Identities: Simplify expressions involving trigonometric functions or prove identities.
• Applications: Real-world problems such as finding the height of an object (e.g., building or tree) using angles of elevation or depression.

6.0Sample NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

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

(i) sin A, cos A

(ii) sin C, cos C.

Solution.

By Pythagoras Theorem,

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

⇒ AC = √625 = 25 cm.

(i) sin A = BC/AC {i.e., (side opposite to angle A) / Hyp.}

= 7/25 (since BC = 7 cm and AC = 25 cm)

cos A = AB/AC {i.e., (side adjacent to angle A) / Hyp.}

= 24/25 (since AB = 24 cm and AC = 25 cm)

(ii) sin C = AB/AC {i.e., (side opposite to angle C) / Hyp.}

= 24/25 

cos C = BC/AC {i.e., (side adjacent to angle C) / Hyp.}

= 7/25

2. In the given figure, find tan P - cot R.

Solution:

In the given figure, by the Pythagoras Theorem,

QR² = PR² - PQ² = (13)² - (12)² = 25

=> QR = √25 = 5 cm

In triangle PQR, which is right-angled at Q, QR = 5 cm is the side opposite to angle P, and PQ = 12 cm is the side adjacent to angle P.

Therefore, tan P = QR/PQ = 5/12.

Now, QR = 5 cm is the side adjacent to angle R, and PQ = 12 cm is the side opposite to angle R.

Therefore, cot R = QR/PQ = 5/12.

Hence, tan P - cot R = 5/12 - 5/12 = 0.

3. Evaluate:

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° - sin² 60°

(iii) cos 45° / (sec 30° + cosec 30°)

(iv) (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°)

(v) (5 cos² 60° + 4 sec² 30° - tan² 45°) / (sin² 30° + cos² 30°)

Sol.

(i) sin 60° cos 30° + sin 30° cos 60°

= (√3 / 2)(√3 / 2) + (1 / 2)(1 / 2) = (√3 / 2)² + (1 / 2)²

= 3 / 4 + 1 / 4 = 1

(ii) 2 tan² 45° + cos² 30° - sin² 60°

= 2 × (1)² + (√3 / 2)² - (√3 / 2)²

= 2 + 3 / 4 - 3 / 4 = 2

(iii) cos 45° / (sec 30° + cosec 30°)

= (1 / √2) / (2 / √3 + 2) = (1 / √2) / (2(1 + √3) / √3) = 1(√3) / (2√2(1 + √3))

= √3 / (2√2) × (√3 - 1) / ((√3 + 1)(√3 - 1)) = √3(√3 - 1) / (2√2 × 2)

= (3 - √3) / (4√2) = (3 - √3) / (4√2) × (√2 / √2) = (3√2 - √6) / 8

(iv) (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°)

= (1 / 2 + 1 - 2 / √3) / (2 / √3 + 1 / 2 + 1)

= ((√3 + 2√3 - 4) / (2√3)) / ((4 + √3 + 2√3) / (2√3)) = (3√3 - 4) / (4 + 3√3) × (4 - 3√3) / (4 - 3√3)

= (12√3 - 27 - 16 + 12√3) / (16 - 9 × 3)

= (24√3 - 43) / (-11) = (43 - 24√3) / 11

(v) (5 cos² 60° + 4 sec² 30° - tan² 45°) / (sin² 30° + cos² 30°)

= (5(cos 60°)² + 4(sec 30°)² - (tan 45°)²) / ((sin 30°)² + (cos 30°)²)

= (5(1 / 2)² + 4(2 / √3)² - (1)²) / ((1 / 2)² + (√3 / 2)²) = (5 / 4 + 4 × 4 / 3 - 1) / (1 / 4 + 3 / 4)

= (5 / 4 + 16 / 3 - 1) / (1 / 4 + 3 / 4) = 5 / 4 + 16 / 3 - 1

= (15 + 64 - 12) / 12 = 67 / 12

4. If tan(A+B) = √3 and tan(A-B) = 1/√3; 0° < A+B ≤ 90°; A > B, find A and B.

Sol. tan(A+B) = √3  => A+B = 60°  ...(1)

tan(A-B) = 1/√3 => A-B = 30°  ...(2)

Adding (1) and (2),

2A = 90° => A = 45°

Then from (1), 45° + B = 60° => B = 15°

5. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

(i) sin A = √(1 - cos² A)

= √(1 - 1/sec² A) = √(sec² A - 1)/sec A

(ii) cos A = 1/sec A

(iii) tan A = √(sec² A - 1)

(iv) cot A = 1/tan A = 1/√(sec² A - 1)

(v) cosec A = 1/sin A = sec A/√(sec² A - 1)

7.0Benefits of Class 10 Maths Chapter 8 Introduction to Trigonometry

The Introduction to Trigonometry chapter in Class 10 Maths offers several key benefits:

1. Foundation for Advanced Trigonometry: It introduces basic trigonometric ratios (sine, cosine, tangent) that are crucial for higher-level studies in mathematics and physics.
2. Real-Life Applications: Trigonometry is widely used in fields like engineering, architecture, navigation, and astronomy to calculate distances, angles, and heights.
3. Problem-Solving Skills: Students develop skills to solve practical problems involving right-angled triangles.
4. Analytical Thinking: The chapter enhances logical thinking and the ability to apply mathematical concepts to real-world situations.
5. Link to Further Topics: It forms the basis for more advanced trigonometric concepts that students will encounter in higher classes.