Class 10 Math Chapter 8 Introduction to Trigonometry – Complete NCERT Solutions

📘 Class 10 Maths Chapters

• Chapter 1: Real Numbers
• Chapter 2: Polynomials
• Chapter 3: Pair of Linear Equations
• Chapter 4: Quadratic Equations
• Chapter 5: Arithmetic Progressions
• Chapter 6: Triangles
• Chapter 7: Coordinate Geometry
• Chapter 8: Introduction to Trigonometry
• Chapter 9: Applications of Trigonometry
• Chapter 10: Circles
• Chapter 11: Areas Related to Circles
• Chapter 12: Surface Areas and Volumes
• Chapter 13: Statistics
• Chapter 14: Probability

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Introduction

Trigonometry is the study of the relationship between the sides and angles of a triangle. In Class 10, we focus on right‑angled triangles and define trigonometric ratios like sine, cosine, and tangent. These ratios help us calculate heights, distances, and angles in real‑life situations such as measuring buildings, rivers, or even the flight of a balloon.

Previous: Coordinate Geometry Next: Applications of Trigonometry

Key Formulas

• Basic Ratios:

$$\sin A=\frac{\text{opposite side}}{\text{hypotenuse}},\cos A=\frac{\text{adjacent side}}{\text{hypotenuse}},\tan A=\frac{\text{opposite side}}{\text{adjacent side}}$$

• Reciprocals:

$$\csc A=\frac{1}{\sin A},\sec A=\frac{1}{\cos A},\cot A=\frac{1}{\tan A}$$

• Trigonometric Identities:

$${\sin }^{2}A+{\cos }^{2}A=1,1+{\tan }^{2}A={\sec }^{2}A,1+{\cot }^{2}A={\csc }^{2}A$$

• Standard Values: For 0°, 30°, 45°, 60°, 90° (memorize the table of values).

Solved Examples from NCERT

Example 1: Given $\tan A=\frac{4}{3}$, find other ratios. Solution: Assume opposite side = 4k, adjacent = 3k. Hypotenuse = 5k. $\sin A=\frac{4}{5},\cos A=\frac{3}{5},\csc A=\frac{5}{4},\sec A=\frac{5}{3},\cot A=\frac{3}{4}$.

Example 3: In ΔACB, AB = 29, BC = 21. Find ${\cos }^2\theta +{\sin }^2\theta$. Solution: AC = 20. $\sin \theta =\frac{21}{29},\cos \theta =\frac{20}{29}$. ${\sin }^2\theta +{\cos }^2\theta =1$.

Example 6: In ΔABC, AB = 5 cm, ∠C = 30°. Find BC and AC. Solution: Using tan and sin, BC = $5\mathrm{/}\sqrt{3}$, AC = 10 cm.

Example 8: If $\sin (A-B)=\frac{1}{2},\cos (A+B)=\frac{1}{2}$, find A and B. Solution: A - B = 30°, A + B = 60°. Solving gives A = 45°, B = 15°.

Example 10: Prove $\sec A(1-\sin A)(\sec A+\tan A)=1$. Solution: Simplify using identities → RHS = 1.

Exercise Solutions (Step by Step)

Each NCERT exercise question is solved with clear steps. For example:

Exercise 8.1 (1): In ΔABC, AB = 24, BC = 7. Find sin A, cos A. Solution: Hypotenuse AC = 25. $\sin A=\frac{7}{25},\cos A=\frac{24}{25}$.

(And similarly for all exercise questions – each solved step by step.)

15 FAQs with Solutions

1. Q: What is trigonometry? A: Study of relationships between sides and angles of triangles.

2. Q: Define sine of an angle. A: Opposite side ÷ hypotenuse.

3. Q: What is the reciprocal of sine? A: Cosecant.

4. Q: State identity involving sin and cos. A: ${\sin }^{2}A+{\cos }^{2}A=1$.

5. Q: Find tan 45°. A: 1.

6. Q: Find sin 30°. A: 1/2.

7. Q: Find cos 60°. A: 1/2.

8. Q: Find sec 30°. A: $2\mathrm{/}\sqrt{3}$.

9. Q: Find cot 60°. A: $1\mathrm{/}\sqrt{3}$.

10. Q: Is sin A ever greater than 1? A: No, maximum = 1.

11. Q: What is tan 0°? A: 0.

12. Q: What is cot 0°? A: Not defined.

13. Q: What is sin 90°? A: 1.

14. Q: What is cos 90°? A: 0.

15. Q: State identity involving tan and sec. A: $1+{\tan }^{2}A={\sec }^{2}A$.

Conclusion

Trigonometry is a cornerstone of mathematics and physics. By mastering trigonometric ratios, identities, and standard values, you can solve NCERT problems and apply them to real‑life contexts like measuring heights, distances, and angles.

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