NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

Class 10 NCERT Solutions Maths Chapter 8 Exercise 8.1 helps students understand the foundational concepts of trigonometry, especially the concepts of trigonometric ratios. The exercise introduces the relationship between the angles and sides of a right-angled triangle. These concepts form a solid base for solving advanced problems in students’ future mathematical studies. The exercise includes some basic questions about foundational trigonometry that can help students gain insightful expertise in the topic. So, let’s start exploring the fundamentals of trigonometry.

1.0Download NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1 : Free PDF

2.0Introduction to Trigonometry:

Before getting into the basic concepts of the chapter, let’s understand the actual meaning of trigonometry itself. The word Trigonometry comes from the Greek words “tri”, which means three; “gon”, meaning sides, and “metron”, meaning measure. It is one of the most important branches of mathematics, astronomy, physics, and architecture. Trigonometry focuses on the relationship between the sides of a right-angled triangle and its angles. These right-angled triangles can be seen in many real-life situations as well. 

For instance, to ensure stability buildings are made at a 90-degree angle to the ground. This relation helps in finding the height and distances of such buildings. In the exercise, we will be looking forward to some basic questions of trigonometry to form a base for the topic. 

3.0Key Concepts: 

The key focus of exercise 8.1 is the study of trigonometric ratios. So, let’s explore some key concepts related to these ratios to solve the questions of this exercise with ease: 

Trigonometric Ratios of a Right-Angled Triangle: 

Trigonometric Ratios of a Right-Angled Triangle express the relationship between angles (particularly acute angles in this exercise) and sides of a right-angled triangle. To understand this better, let’s take an example of a triangle ABC with AB as the base and BC perpendicular. In the triangle, consider angle B equal to 90 degrees and ∠CAB as an acute angle. Here, understand the following terms of this triangle: 

• Opposite side: The side opposite to the angle of interest, which means the perpendicular. 
• Adjacent side: The side next to the angle, but not the hypotenuse, meaning base.
• Hypotenuse: The longest side, opposite the right angle. 

Trigonometric Ratios for the Angle A: 

Based on the sides of the above-mentioned triangle in the example, there will be a total of six trigonometric ratios for angle A. These ratios are: 

• Sine (sin A): The ratio of the length of the opposite side to the hypotenuse.

$SinA=\frac{SideOppositetoangleA}{Hypotenuse}=\frac{Perpendicular}{Hypotenuse}$

• Cosine (cos A): The ratio of the length of the adjacent side to the hypotenuse.

$CosA=\frac{SideadjacenttoangleA}{Hypotenuse}=\frac{Base}{Hypotenuse}$

• Tangent (tan A): The ratio of the length of the opposite side to the adjacent side.

$tanA=\frac{SideOppositetoangleA}{SideadjacenttoangleA}=\frac{Perpendicular}{Base}$

• Cosecant (cosec A): The reciprocal of sine, i.e., the ratio of the hypotenuse to the opposite side.

$cosecA=\frac{Hypotenuse}{SideOppositetoangleA}=\frac{Hypotenuse}{Perpendicular}$

• Secant (sec A): The inverse of cosine, i.e., the hypotenuse to the adjacent side ratio.

$secA=\frac{Hypotenuse}{SideadjacenttoangleA}=\frac{Hypotenuse}{Base}$

• Cotangent (cot A): The inverse of a tangent, i.e., the ratio of the adjacent side to the opposite side.

$cotA=\frac{SideAdjacenttoangleA}{SideOppositetoangleA}=\frac{Base}{Perpendicular}$

Remark: 

• As the hypotenuse is the longest side in a right-angled triangle, the value of sinA or cosA always remains less than or equal to 1. 
• The trigonometric ratios remain constant for the same angle across different right-angled triangles, regardless of their size.

Trigonometric Ratios for Other Angles

As per the formulas for each trigonometric ratio, we can clearly see that each ratio is connected directly or indirectly with other angles' ratios. As we proceed in the exercise, we will calculate the other angles by just one trigonometric ratio using these relations and without the need for complex calculations. 

Start solving the problems with the NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1 to master these concepts and apply them effectively in different mathematical and real-life situations, s

4.0NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1 : Detailed Solution

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

Solution:

In the given figure,

sin A = 3/4

=> BC/AC = 3/4

=> BC = 3k

and AC = 4k

where k is the constant of proportionality.

By Pythagoras Theorem,

AB² = AC² - BC² = (4k)² - (3k)² = 7k²

=> AB = √(7k²) = k√7

So, cos A = AB/AC = (k√7) / (4k) = √7 / 4

and tan A = BC/AB = (3k) / (k√7) = 3 / √7

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

Solutions:

cot A = 8/15

=> AB/BC = 8/15

=> AB = 8k and BC = 15k

Now, AC = √((8k)² + (15k)²) = √(64k² + 225k²) = √(289k²) = 17k

sin A = BC/AC = 15k/17k = 15/17

sec A = AC/AB = 17k/8k = 17/8

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

Sol.

Given: sec θ = 13/12

This implies AC/BC = 13/12, where AC is the hypotenuse and BC is the adjacent side of the right-angled triangle.

Using Pythagoras Theorem:

AC² = AB² + BC²

(13k)² = AB² + (12k)²  (where k is a positive constant)

AB² = 169k² - 144k²

AB² = 25k²

AB = √(25k²) = 5k

Now, we can find the other trigonometric ratios:

sin θ = AB/AC = (5k)/(13k) = 5/13

cos θ = BC/AC = (12k)/(13k) = 12/13

tan θ = AB/BC = (5k)/(12k) = 5/12

cot θ = BC/AB = (12k)/(5k) = 12/5

cosec θ = AC/AB = (13k)/(5k) = 13/5

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

Solution:

Let's consider a triangle ABC where ∠A and ∠B are acute angles.

Draw a CD perpendicular to AB.

Given: cos A = cos B

We need to show that ∠A = ∠B.

=> AD/AC = BD/BC = k ... (1)

AD = kAC

BD = kBC

In triangle ACD,

CD = √(AC² - AD²)

CD = √(AC² - k²AC²)

CD = AC√(1 - k²)

In triangle BCD,

CD = √(BC² - BD²)

CD = √(BC² - k²BC²)

CD = BC√(1 - k²)

From equations (2) and (3),

CD/CD = AC/BC

AC = BC

=> ∠A = ∠B (Angles opposite to equal sides are equal)

7. If cot θ = 7/8, evaluate:

(i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ)

(ii) cot² θ

Sol. In the figure,

cot θ = 7/8

=> AB/BC = 7/8

=> AB = 7k and BC = 8k

Now, AC² = AB² + BC² = (7k)² + (8k)² = 113k²

=> AC = √(113)k

Then sin θ = BC/AC = 8k/√(113)k = 8/√113

and cos θ = AB/AC = 7k/√(113)k = 7/√113.

(i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ) = 

(1 + 8/√113)(1 - 8/√113) / (1 + 7/√113)(1 - 7/√113) = 

(√113 + 8)(√113 - 8) / (√113 + 7)(√113 - 7) = 

((√113)² - (8)²) / ((√113)² - (7)²)  {∵ (a + b)(a - b) = a² - b²} = 

(113 - 64) / (113 - 49) = 49/64

(ii) cot² θ = (7/8)² = 49/64

8. If 3 cot A = 4, check whether (1 - tan² A) / (1 + tan² A) = cos² A - sin² A or not.

Sol.

3 cot A = 4

=> cot A = 4/3

=> AB/BC = 4/3

=> AB = 4k and BC = 3k

Now, AC = √((4k)² + (3k)²) = 5k

Then sin A = BC/AC = 3k/5k = 3/5, 

cos A = AB/AC = 4k/5k = 4/5, 

and tan A = BC/AB = 3k/4k = 3/4

LHS = (1 - tan² A) / (1 + tan² A) = (1 - (3/4)²) / (1 + (3/4)²) = 

(1 - 9/16) / (1 + 9/16) = (16 - 9) / (16 + 9) = 7/25

RHS = cos² A - sin² A = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25

Therefore, LHS = RHS, i.e., (1 - tan² A) / (1 + tan² A) = cos² A - sin² A  (∵ Each side = 7/25)

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.

Sol.

tan A = 1/√3

BC/BA = 1/√3

BC = k and BA = √3 k

AC² = BC² + BA²

= k² + (√3 k)² = k² + 3k² = 4k²

AC = √(4k²) = 2k

(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) = √3/4 - √3/4 = 0

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

Sol. In figure,

P Q = 5 cm

PR + QR = 25 cm

i.e., PR = 25 cm - QR

Now, PR² = PQ² + QR²

=> (25 - QR)² = (5)² + QR²

=> 625 - 50 × QR + QR² = 25 + QR²

=> 50 × QR = 600

=> QR = 12 cm

and PR = 25 cm - 12 cm = 13 cm

We find sin P = QR / PR = 12 / 13, cos P = PQ / PR = 5 / 13 and 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.

Sol.

(i) False, We know that tan 60° = √3 > 1.

(ii) True, We know that the value of sec A is always ≥ 1. Since 12/5 is greater than one, this statement is true.

(iii) False, Because cos A is the abbreviation used for cosine A.

(iv) False, because cot A is not the product of cot and A. cot A represents the cotangent of angle A.

(v) False, because the value of sin θ cannot be more than 1.

5.0Benefits of Class 10 Maths Chapter 8 Introduction to Trigonometry

• Builds understanding of trigonometric ratios (sin, cos, tan).
• Helps solve right-angled triangle problems.
• Improves logical and analytical thinking.
• Establishes key concepts for future studies in physics and engineering.
• Lays the foundation for more advanced trigonometry topics and competitive exams.