NCERT Solutions for Class 10 Maths Chapter 8 PDF Download
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in Chapter 8.
Ananya Gupta19 Mar, 2024
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NCERT Solutions for Class 10 Maths Chapter 8: NCERT Solutions for Class 10 Maths Chapter 8, Introduction to Trigonometry, are helpful resources for students. They make it easier to understand tricky concepts and score well in the CBSE Class 10 board exam.
These solutions are created by subject experts and cover all the questions from the textbook. These solutions are based on the latest CBSE Syllabus for 2023-24 and match the exam pattern ensuring students are well-prepared.NCERT Solutions for Class 10 Maths Chapter 8 PDF
You can access the NCERT Solutions for Class 10 Maths Chapter 8 through the provided PDF link. These solutions provide detailed explanations and step-by-step solutions to all the exercises in the chapter.NCERT Solutions for Class 10 Maths Chapter 8 PDF
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Exercise 8.1
1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos CAnswer:
Let us draw a right angled triangle ABC, right angled at B. Using Pythagoras theorem,
(i)
(ii)
,
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2. In 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
B =90
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/√7
4. 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.
Answer:
Consider a triangle ABC in which
Let AB = 12k and AC = 13k
Then, using Pythagoras theorem,
(AC)2 = (AB)2 + (BC)2
(13
k
)2 = (12
k
)2 + (BC)2
169
k
2 = 144
k
2 + BC2
25
k
2 = BC2
BC = 5
k
(I) Sin
=
=
=
(ii) Cos θ =
=
=
(iii)Tan θ =
=
=
(iv) Cot θ =
=
=
(v) Cosec θ =
=
=
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
cos A = cos But
7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot
2
θ
Answer:
Consider a triangle ABC
(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.
Answer:
Consider a triangle ABC AB=4cm, BC= 3cm
.
And
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)
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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. 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, Hypotenuse is the longest side. ∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.Exercise 8.2
1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan 2 45° + cos 2 30° – sin 2 60°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos 2 60° + 4sec 2 30° - tan 2 45°)/(sin 2 30° + cos 2 30°)
Answer:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan
2
45° + cos
2
30° – sin
2
60°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos
2
60° + 4sec
2
30° - tan
2
45°)/(sin
2
30° + cos
2
30°)
2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan
2
30° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan
2
45°/1+tan
2
45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin
2
A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan
2
30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Answer:
(i) 2tan 30°/1+tan 2 30° = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° =
(ii) 1-tan
2
45°/1+tan
2
45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin
2
A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan
2
30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
Answer:
……….(i)
4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Answer:
(i) False. Let A = 30° and B = 60°, then sin (A + B) = sin (30° + 60°) = sin 90° = 1 and, sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = 1+√3/2 (ii) True. sin 0° = 0 sin 30° = 1/2 sin 45° = 1/√2 sin 60° = √3/2 sin 90° = 1 Thus the value of sin θ increases as θ increases. (iii) False. cos 0° = 1 cos 30° = √3/2 cos 45° = 1/√2 cos 60° = 1/2 cos 90° = 0 Thus the value of cos θ decreases as θ increases. (iv) True. cot A = cos A/sin A cot 0° = cos 0°/sin 0° = 1/0 = undefined.Exercise 8.3
1. Evaluate : (i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° – sin 42° (iv) cosec 31° – sec 59°Answer:
(i) sin 18°/cos 72° = sin (90° - 18°) /cos 72° = cos 72° /cos 72° = 1 (ii) tan 26°/cot 64° = tan (90° - 36°)/cot 64° = cot 64°/cot 64° = 1 (iii) cos 48° - sin 42° = cos (90° - 42°) - sin 42° = sin 42° - sin 42° = 0 (iv) cosec 31° - sec 59° = cosec (90° - 59°) - sec 59° = sec 59° - sec 59° = 0 2. Show that : (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° – sin 38° sin 52° = 0Answer:
(i) tan 48° tan 23° tan 42° tan 67° = tan (90° - 42°) tan (90° - 67°) tan 42° tan 67° = cot 42° cot 67° tan 42° tan 67° = (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1 (ii) cos 38° cos 52° - sin 38° sin 52° = cos (90° - 52°) cos (90°-38°) - sin 38° sin 52° = sin 52° sin 38° - sin 38° sin 52° = 0 3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.Answer:
tan 2A = cot (A- 18°) ⇒ cot (90° - 2A) = cot (A -18°) Equating angles, ⇒ 90° - 2A = A- 18° ⇒ 108° = 3A ⇒ A = 36° 4. If tan A = cot B, prove that A + B = 90°.Answer:
tan A = cot B ⇒ tan A = tan (90° - B) ⇒ A = 90° - B ⇒ A + B = 90° 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.Answer:
sec 4A = cosec (A - 20°) ⇒ cosec (90° - 4A) = cosec (A - 20°) Equating angles, 90° - 4A= A- 20° ⇒ 110° = 5A ⇒ A = 22° 6. If A, B and C are interior angles of a triangle ABC, then show that sin (B+C/2) = cos A/2Answer:
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°) = cos 23° + sin 15°Exercise 8.4
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer:
3. Evaluate :
(i) (sin
2
63° + sin
2
27°)/(cos
2
17° + cos
2
73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Answer:
(i) (sin 2 63° + sin 2 27°)/(cos 2 17° + cos 2 73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
4. (i) 9 sec
2
A - 9 tan
2
A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (secA + tanA) (1 - sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan
2
A/1+cot
2
A =
(A) sec
2
A (B) -1 (C) cot
2
A (D) tan
2
A
Answer:
(i) (i) 9 sec 2 A - 9 tan 2 A = (A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (secA + tanA) (1 - sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan
2
A/1+cot
2
A =
(A) sec
2
A (B) -1 (C) cot
2
A (D) tan
2
A
5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ - cot θ)
2
= (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin
2
A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec
2
A = 1+cot
2
A.
(vi) √1 + sin A/1 - sin A = sec A+ tan A
(vii) (sin θ - 2sin
3
θ)/(2cos
3
θ-cos θ) = tan θ
(viii) (sin A + cosec A)
2
+ (cos A + sec A)
2
= 7+tan
2
A+cot
2
A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan
2
A/1+cot
2
A) = (1-tan A/1-cot A)
2
= tan
2
A
Answer:
(i) (cosec θ - cot θ) 2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin
2
A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec
2
A = 1+cot
2
A
(vi) √1 + sin A/1 - sin A = sec A+ tan A
LHS = 1 + sin A/(1 - sin A) .....(1)
Multiplying and dividing by (1 + sin A)
⇒ (1 + sin A)(1 + sin A/1 - sin A)(1 + sin A)
= (1 + sin A)²/(1 - sin² A) [a² - b² = (a - b)(a + b)]
= (1 + sinA)/1 - sin² A
= 1 + sin A/cos² A
= 1 + sin A/cos A
= 1/cos A + sin A/cos A
= sec A + tan A
= R.H.S
(vii) (sin θ - 2sin
3
θ)/(2cos
3
θ-cos θ) = tan θ
(viii) (sin A + cosec A)
2
+ (cos A + sec A)
2
= 7+tan
2
A+cot
2
A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan
2
A/1+cot
2
A) = (1-tan A/1-cot A)
2
= tan
2
A
NCERT Solutions for Class 10 Maths Chapter 8 FAQs
What is trigonometry?
Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles, particularly right triangles.
What are trigonometric ratios?
Trigonometric ratios are ratios of the lengths of two sides of a right triangle. The main trigonometric ratios are sine, cosine, and tangent, which are abbreviated as sin, cos, and tan respectively.
How are trigonometric ratios calculated?
Trigonometric ratios are calculated using the lengths of the sides of a right triangle. For example, the sine of an angle in a right triangle is calculated as the length of the side opposite the angle divided by the length of the hypotenuse.
What are trigonometric identities?
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables. Examples include the Pythagorean identity, sum and difference identities, double angle identities, and half angle identities.
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