Class 10 Maths Chapter 9 Introduction to Trigonometry Exercise 9.1 NCERT Solutions
Ananya Gupta9 May, 2026
NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1 introduce the basic trigonometric ratios: sine, cosine, and tangent. This exercise will help you understand how these ratios are defined in a right-angled triangle based on the relationship between its sides, as outlined in the CBSE Class 10 syllabus.
These NCERT solutions are explained step by step to make it easier to identify the sides correctly and apply the ratios accurately. Practising these questions builds a strong foundation for solving trigonometric problems in later exercises.
NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1
1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Answer:
Let us draw a right-angled triangle ABC, right-angled at B. Using Pythagoras' theorem
(ii)
,2. In the adjoining figure, find tan P – cot R.
Answer:
3. If sin A =3/4, calculate cos A and tan A.
Answer:
Given: A triangle ABC in which
We know that sin A = BC/AC = 3/4.
Let BC be 3k, and AC will be 4k, where k is a positive real number.
By Pythagoras theorem we get, AC 2 = AB 2 + BC 2 (4k) 2 = AB 2 + (3k) 2 16k 2 - 9k 2
= AB 2 AB 2 = 7k 2 AB = √7 k cos A = AB/AC = √7 k/4k = √7/4 tan A = BC/AB = 3k/√7 k = 3/√74. Given 15 cot A = 8, find sin A and sec A.
Answer:
Let ΔABC be a right-angled triangle, right-angled at B. We know that cot A = AB/BC = 8/15
(Given). Let AB be 8k, and BC will be 15k, where k is a positive real number.
By Pythagoras theorem we get, AC 2 = AB 2 + BC 2 AC 2 = (8k) 2 + (15k) 2
AC 2 = 64k 2 + 225k 2
AC 2 = 289k 2 AC = 17 k sin A = BC/AC = 15k/17k = 15/17 sec A = AC/AB = 17k/8 k = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Consider a triangle ABC in which
We know that the sec function is the reciprocal of the cos function, which is equal to the ratio of the length of the hypotenuse side to the adjacent side
Let us assume a right-angled triangle ABC, right-angled at B
sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB
Let AC be 13k and AB will be 12k
Where k is a positive real number.
According to the Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle, and we get,
AC2=AB2 + BC2
Substitute the value of AB and AC
(13k)2= (12k)2 + BC2
169k2= 144k2 + BC2
169k2= 144k2 + BC2
BC2 = 169k2 – 144k2
BC2= 25k2
Therefore, BC = 5k
Now, substitute the corresponding values in all other trigonometric ratios
So,
Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13
Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13
tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12
Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5
cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot 2 θAnswer:
(ii)
8. If 3cot A = 4/3 , check whether (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A or not.
Consider a triangle ABC AB=4cm, BC= 3cm
.
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
Answer:
Consider a triangle ABC in which
.
(i)
(ii)
10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Answer:
Given that, PR + QR = 25 , PQ = 5. Let PR be x.
∴ QR = 25 - x
By Pythagoras theorem , PR2 = PQ 2 + QR 2 x 2 = (5)2 + (25 - x) 2 x 2 = 25 + 625 + x 2 - 50x 50x = 650 x = 13
∴ PR = 13 cm QR = (25 - 13) cm = 12 cm sin P = QR/PR = 12/13 cos P = PQ/PR = 5/13 tan P = QR/PQ = 12/5
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 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 = 4/3 for some angle θ.
Answer:
i) False. In ΔABC in which ∠B = 90º, AB = 3, BC = 4 and AC = 5 Value of tan A = 4/3 which is greater than. The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem. AC 2 = AB 2 + BC 2 5 2 = 3 2 + 4 2 25 = 9 + 16 25 = 25
(ii) True. Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number. By Pythagoras' theorem, we get, AC 2 = AB 2 + BC 2 (12k) 2 = (5k) 2 + BC 2 BC 2 + 25k 2 = 144k 2 BC 2 = 119k 2 Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False. The abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.
(iv) False. cot A is not the product of cot and A. It is the cotangent of ∠A. (v) False. sin θ = Height/Hypotenuse. We know that in a right-angled triangle, the Hypotenuse is the longest side. ∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.
How to Score Better in Class 10 Maths Exam?
Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:
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Build Strong Concepts:
Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.
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Work on Weak Areas:
Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.
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Revise Formulas Daily:
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Practise Regularly:
Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.
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Solve Previous Year Papers:
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