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.
Students can download NCERT Solutions pdf for class 10 Maths Chapter 8 from the link below:
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.
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 . These concepts provide a foundation for solving real-life problems involving angles and distances.
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:
Standard Angles:
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.
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 to find the angle.
Step 6: Verify the solution (check if the side lengths satisfy the triangle’s properties or trigonometric identities).
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)
The Introduction to Trigonometry chapter in Class 10 Maths offers several key benefits:
(Session 2025 - 26)