Introduction of Trigonometry

August 23, 2024

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NCERT Solutions for Class 10 Maths Chapter 8

The word “trigonometry” is derived from the Greek words “tri” (meaning three), “gon” (meaning sides), and “metron” (meaning measure).

Key Concepts:

Trigonometric Ratios: These are ratios of the sides of a right-angled triangle with respect to its acute angles.
Trigonometric Identities: These are equations involving trigonometric ratios that hold true for all values of the angle involved.
Trigonometric Tables: These tables provide the values of trigonometric ratios for different angles.

Applications of Trigonometry:

Trigonometry has wide-ranging applications in various fields, including:

Surveying
Navigation
Engineering
Physics
Astronomy

In essence, trigonometry provides a mathematical framework for understanding and analyzing geometric relationships, especially in triangles.

NCERT Solutions for Class 10 Maths Chapter 8

Exercise 8.1

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

(i) sin A, cos A

(ii) sin C, cos C

Ans :

Step 1: Find the hypotenuse AC Using the Pythagorean theorem: AC² = AB² + BC² AC² = 24² + 7² AC² = 576 + 49 AC² = 625 AC = 25 cm

Step 2: Find the trigonometric ratios (i) sin A, cos A

sin A = opposite side / hypotenuse = BC / AC = 7/25
cos A = adjacent side / hypotenuse = AB / AC = 24/25

(ii) sin C, cos C

sin C = opposite side / hypotenuse
= AB / AC = 24/25
cos C = adjacent side / hypotenuse
= BC / AC = 7/25

2. In given figure, find tan P – cot R.

Ans :

3. If sin A = 3/4 , calculate cos A and tan A.

Ans :

Using the Pythagorean Identity:

sin²A + cos²A = 1

Substituting the given value of sin A:

(3/4)² + cos²A = 1
9/16 + cos²A = 1
cos²A = 1 – 9/16
cos²A = 7/16
cos A = ±√(7/16)

Since we are dealing with an angle in a right-angled triangle, cosine is positive. Therefore, cos A = √7/4

Finding tan A:

tan A = sin A / cos A = (3/4) / (√7/4) = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Ans :

Given: 15 cot A = 8 Therefore, cot A = 8/15

We know that:

cot A = base/perpendicular = B/P
So, let base (B) = 8k and perpendicular (P) = 15k, where k is a positive constant.

Using Pythagoras theorem:

Hypotenuse (H)² = Base² + Perpendicular²
H² = (8k)² + (15k)²
H² = 64k² + 225k²
H² = 289k²
H = 17k

Find sin A and sec A:

sin A = perpendicular/hypotenuse = P/H = 15k/17k = 15/17
sec A = hypotenuse/base = H/B = 17k/8k = 17/8

NCERT Solutions for Class 10 Maths Chapter 8

5. Given sec θ = 13/12 , calculate all other trigonometric ratios.

Ans :

Given: sec θ = 13/12

We know that:

sec θ = hypotenuse / base

Therefore, hypotenuse = 13 and base = 12.

Using Pythagoras theorem to find the perpendicular:

hypotenuse² = perpendicular² + base²
13² = perpendicular² + 12²
perpendicular² = 169 – 144 = 25
perpendicular = 5

Now, we can find the other trigonometric ratios:

sin θ = perpendicular / hypotenuse
= 5/13
cos θ = base / hypotenuse = 12/13
tan θ = perpendicular / base = 5/12
cot θ = base / perpendicular = 12/5
cosec θ = hypotenuse / perpendicular = 13/5

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Ans :

Consider a right-angled triangle ABC with ∠C = 90°.

cos A = BC/AC (adjacent side/hypotenuse)
cos B = AC/AB (adjacent side/hypotenuse)

Given that cos A = cos B, we have:

BC/AC = AC/AB

Cross-multiplying, we get:

BC * AB = AC * AC
BC * AB = AC²

Now, consider the same triangle ABC.

By the Pythagorean theorem, we have:
- AB² = AC² + BC²

Substitute BC * AB for AC² from the previous equation:

AB² = BC * AB + BC²
AB² – BC * AB – BC² = 0
(AB – BC)(AB + BC) = 0

Since AB and BC are lengths of sides of a triangle, they cannot be negative. Therefore, AB + BC ≠ 0.

Hence, AB – BC = 0

AB = BC

In triangle ABC, since AB = BC, the angles opposite to these equal sides are also equal.

Therefore, ∠A = ∠B.

NCERT Solutions for Class 10 Maths Chapter 8

7. If cot θ = 7/8, evaluate:

(i) (1+sinθ)(1−sinθ)/(1+cosθ)(1−cosθ)

(ii) cot²θ

Ans :

(i) Evaluating (1+sinθ)(1-sinθ) / (1+cosθ)(1-cosθ)

Using the identity (a+b)(a-b) = a²-b², we get:

Numerator = (1+sinθ)(1-sinθ) = 1 – sin²θ
Denominator = (1+cosθ)(1-cosθ) = 1 – cos²θ

Using the Pythagorean identity sin²θ + cos²θ = 1, we get:

Numerator = cos²θ
Denominator = sin²θ

Therefore, the expression becomes:

cos²θ / sin²θ = (cosθ/sinθ)²

Since cot θ = cosθ/sinθ, we have:

(cosθ/sinθ)² = cot²θ

Now, we know cot θ = 7/8, so cot²θ = (7/8)² = 49/64.

Therefore, the value of the expression is 49/64.

8. If 3 cot A = 4, check whether 1−tan2A/1+tan2A = cos² A – sin² A or not. NCERT Solutions for Class 10 Maths Chapter 8

Ans :

Given: 3 cot A = 4

Step 1: Find tan A

cot A = 4/3
tan A = 1/cot A = 3/4

Step 2: Evaluate the Left Hand Side (LHS)

LHS = (1 – tan²A) / (1 + tan²A) = (1 – (3/4)²) / (1 + (3/4)²) = (1 – 9/16) / (1 + 9/16) = (7/16) / (25/16) = 7/25

Step 3: Evaluate the Right Hand Side (RHS)

RHS = cos²A – sin²A

To find cos A and sin A, we can use the Pythagorean identity:

tan²A + 1 = sec²A
(3/4)² + 1 = sec²A
sec²A = 25/16
sec A = 5/4

Now, cos A = 1/sec A = 4/5 And sin A = √(1 – cos²A) = √(1 – (16/25)) = 3/5

Therefore, RHS = (4/5)² – (3/5)² = 16/25 – 9/25 = 7/25

Since LHS = RHS, the given equation is true.

Hence, (1 – tan²A) / (1 + tan²A) = cos²A – sin²A.

NCERT Solutions for Class 10 Maths Chapter 8

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

Ans :

Step 1: Find sin A and cos A

Given tan A = 1/√3

We know that tan A = perpendicular/base = BC/AB

So, let BC = k and AB = √3k

Using Pythagoras theorem, AC = 2k

Therefore, sin A = BC/AC = k/2k = 1/2

And cos A = AB/AC = √3k/2k = √3/2

Step 2: Find the values of sin C and cos C

Since ∠B = 90°, ∠A + ∠C = 90°

So, if tan A = 1/√3, then ∠A = 30°

Therefore, ∠C = 60°

sin C = sin 60° = √3/2

cos C = cos 60° = 1/2

Step 3: Evaluate the expressions

(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1

(ii) cos A cos C – sin A sin C = (√3/2)(1/2) – (1/2)(√3/2) = 0

10. In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Ans :

Step 1: Find the lengths of PR and QR Let QR = x cm. Then, PR = (25 – x) cm.

Using the Pythagorean theorem in right-angled triangle PQR:

PQ² + QR² = PR²
5² + x² = (25 – x)²
25 + x² = 625 – 50x + x²
50x = 600
x = 12

So, QR = 12 cm and PR = 25 – 12 = 13 cm.

Step 2: Find the trigonometric ratios

sin P = opposite side / hypotenuse = QR/PR = 12/13
cos P = adjacent side / hypotenuse = PQ/PR = 5/13
tan P = opposite side / adjacent side
= QR/PQ = 12/5

11. State whether the following statements 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.

Ans :

(i) False.

The value of tan A can be less than, greater than, or equal to 1. It depends on the values of the opposite side and adjacent side of the right-angled triangle.

(ii) True.

sec A is the reciprocal of cos A. Since cos A can take values between -1 and 1 (excluding 0), sec A can be greater than 1. Hence, sec A = 12/5 is possible for some angle A.

(iii) False.

cos A is the abbreviation for cosine of angle A, while cosec A is the abbreviation for cosecant of angle A.

(iv) False.

cot A is the cotangent of angle A, not the product of cot and A.

(v) False.

The value of sin θ always lies between -1 and 1. Since 4/3 is greater than 1, it cannot be the value of sin θ for any angle θ.

NCERT Solutions for Class 10 Maths Chapter 8

Exercise 8.2

1. Evaluate the following:

NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 Q1

Ans :

Expression (i):

sin 60° cos 30° + sin 30° cos 60°

= (√3/2 * √3/2) + (1/2 * 1/2)

= 3/4 + 1/4

= 1

Expression (ii):

2 tan² 45° + cos² 30° – sin² 60°

= 2 * (1)² + (√3/2)² – (√3/2)²

= 2 + 3/4 – 3/4

= 2

Expression (iii):

cos 45° / (sec 30° + cosec 30°)

= (1/√2) / (2/√3 + 2)

= (1/√2) / (2(1 + √3)/√3)

= √3 / (2√2 * (1 + √3))

Rationalizing the denominator:

= √3 / (2√2 * (1 + √3)) * (√2 – √6) / (√2 – √6)

= (√6 – √18) / (4 * (1 – 3))

= (√6 – 3√2) / (-8)

= (3√2 – √6) / 8

Expression (iv):

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

= (1/2 + 1 – 2) / (2/√3 + 1/2 + 1)

= -1/2 / (2/√3 + 3/2)

= -1/2 / ((4 + 3√3)/2√3)

= -√3 / (4 + 3√3)

Rationalizing the denominator:

= -√3 / (4 + 3√3) * (4 – 3√3) / (4 – 3√3)

= (-4√3 + 9) / (16 – 27)

= (9 – 4√3) / (-11)

Expression (v):

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

= (5 * (1/2)² + 4 * (2/√3)² – (1)²) / ((1/2)² + (√3/2)²)

= (5/4 + 16/3 – 1) / (1/4 + 3/4)

= (15 + 64 – 12) / 12

= 67 / 12

2. Choose the correct option and justify your choice:

NCERT Solutions For Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 Q2

Ans :

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

Ans :

Step 1: Identify the angles from the given tangent values

We know that:

tan 60° = √3
tan 30° = 1/√3

Therefore,

tan(A+B) = tan 60°
tan(A-B) = tan 30°

Step 2: Equate the angles From the above equations, we can deduce that:

A + B = 60° (Equation 1)
A – B = 30° (Equation 2)

Step 3: Solve the equations Adding equations (1) and (2):

2A = 90°
A = 45°

Substituting A = 45° in equation (1):

45° + B = 60°
B = 15°

4. State whether the following statements 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°.

Ans :