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NCERT Solutions for Class 10 Maths Chapter 8 : Introduction to Trigonometry

Q: Q1. What are the most important topics in NCERT Class 10 Chapter 8 Introduction to Trigonometry?

Answer: Chapter 8 of Class 10 Maths is mostly concerned with trigonometry, which is a critical topic in Class 10. It introduces ratios and identities, as well as trigonometric ratios of some specific angles, ratios of some complementary angles, and trigonometric identities for solving equations. This chapter must be completed methodically and with great focus by students.

Q: Q2. How many exercises are there in Chapter 8 of Class 10 Maths?

Answer: Chapter 8 of Class 10 Maths Introduction to Trigonometry has 27 questions, 10 of which are basic, 10 of which are moderate, and 7 of which are extended answer type problems. These questions are divided into four exercises.

Q: Q3. What is the importance of Trigonometry?

Answer: The most successful and progressive use of trigonometry is the analysis and simplification of equations utilizing various trigonometric functions such as sine, cosine, tangent, and so on. The analytical application of trigonometry is critical in engineering domains such as mechanical engineering, electronics, and mechatronics. Trigonometry can also be used to roof a house, make the roof inclined (in the case of single-family bungalows), and determine the height of the roof in buildings, among other things. It is employed in the marine and aviation sectors. It is employed in cartography (creation of maps). Trigonometry is used in satellite systems as well.

Trigonometry Class 10 Maths NCERT Solutions Chapter 8 is concerned with the understanding of various aspects of trigonometry. Students preparing for introduction to trigonometry class 10 will be able to clear all their concepts on trigonometry at the root level. Trigonometry is used to refine complicated problems in many physical sciences and engineering topics today. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

NCERT Solutions Introduction to Trigonometry, the eighth chapter of the section, concentrates on the essential concepts such as the trigonometric ratios, trigonometric identities, particular angle trigonometric ratios, and complementary angle trigonometric ratios. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Introduction to Trigonometry, Chapter 8 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The NCERT Solutions for Class 10 Maths chapter 8 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Introduction

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Access NCERT Solutions for Class 10 Maths Chapter 8 : Introduction to Trigonometry

Exercise 8.1

Question 1

In a

Δ

ABC, right angled at B,

A B = 24

cm,

B C = 7

cm. Determine
(i)

sin A, cos A

(ii)

sin C, cos C

Medium

Solution

Verified by Toppr

R.E.F image

Δ

ABC, B is at right angle.

Given, AB=24cm

BC=7cm

using Pythagoras theorem

A B^{2} + B C^{2} = A C^{2}

\Rightarrow (24)^{2} + (7)^{2} = A C^{2}

\Rightarrow A C = (2 4^{2}) + (7)^{2} = 576 + 49 = 625

\Rightarrow A C = 25 C M

(i)sin A=

\frac{B C}{A C}

= \frac{7}{2 5}

cos A=

\frac{A B}{A C}

= \frac{2 4}{2 5}

(ii)sin C=

\frac{A B}{A C}

= \frac{2 4}{2 5}

cos c=

\frac{B C}{A C}

\frac{7}{2 5}

Question 2

In Fig., find

tan P - cot R

Medium

Solution

Verified by Toppr

△ P Q R

P R^{2} = P Q^{2} + Q R^{2}

....(Pythagoras theorem)

1 3^{2} = 1 2^{2} + Q R^{2}

Q R^{2} = 169 - 144 = 25 = 5

tan P = \frac{Q R}{P Q} = \frac{5}{1 2}, cot R = \frac{Q R}{P Q} = \frac{5}{1 2}

∴ tan P - cot R = \frac{5}{1 2} - \frac{5}{1 2} = 0

Question 3

sin A = \frac{3}{4}

, calculate

cos A

and

tan A

cos A = \frac{3}{4} tan A = \frac{2}{5}

cos A = \frac{7}{4} tan A = \frac{3}{7}

cos A = \frac{3}{2} tan A = \frac{2}{5}

cos A = \frac{5}{7} tan A = \frac{7}{3}

Medium

Solution

Verified by Toppr

Given,

sin A = \frac{3}{4}

\Rightarrow \frac{B C}{D C} = \frac{3}{4}

\Rightarrow B C = 3 k

and

A C = 4 k

where

k

is the constant of proportionality.
By Pythagoras theorem, we have

A B^{2} = A C^{2} - B C^{2} = (4 k)^{2} - (3 k)^{2} = 7 k^{2}

\Rightarrow A B = 7 k

So,

cos A = \frac{A B}{A C} = \frac{7 k}{4 k} = \frac{7}{4}

And

tan A = \frac{B C}{A B} = \frac{3 k}{7 k} = \frac{3}{7}

Question 4

Given

15 cot A = 8

, find

sin A

and

sec A .

Medium

Solution

Verified by Toppr

Given,

15 cot A = 8

cot A = \frac{8}{1 5}

= >

tan A = \frac{1 5}{8}

-------(

tan A = \frac{1}{cot A}

)

We know that,

tan θ = \frac{o p p o s i t e S i d e}{a d j a c e n t S i d e}

Consider the attached figure, triangle

A B C

From Pythagoras theorem,

A C^{2} = A B^{2} + B C^{2}

A C^{2} = 8^{2} + 1 5^{2} = 64 + 225 = 289

A C = 17

cos A = \frac{a d j a c e n t S i d e}{H y p o t e n u s e} = \frac{A B}{A C} = \frac{8}{1 7}

sec A = \frac{1}{cos A} = \frac{1}{\frac{8}{1 7}} \frac{1 7}{8}

sin A = \frac{o p p o s i t e S i d e}{H y p o t e n u s e} = \frac{B C}{A C} = \frac{1 5}{1 7}

Question 5

Given

sec θ = \frac{1 3}{1 2}

, calculate all other trignometric ratios.

Hard

Solution

Verified by Toppr

Let

△ A B C

be right angled triangle (right angled at

B

)

sec Q = \frac{A C}{A B} = \frac{1 3}{1 2}

Let

A C = 13 x

and

A B = 12 x

A C^{2} = A B^{2} + B C^{2}

B C^{2} = A C^{2} - A B^{2}

x^{2} = 169 - 144

x^{2} = 25

x = 25 = 5

sin Q = \frac{B C}{A C} = \frac{5 x}{1 3 x} = \frac{5}{1 3}

cos Q = \frac{A B}{A C} = \frac{1 2}{1 3}

tan Q = \frac{B C}{A B} = \frac{5 x}{1 2 x} = \frac{5}{1 2}

cot Q = \frac{1}{tan Q} = \frac{1}{\frac{5}{1 2}} = \frac{1 2}{5}

cosec Q = \frac{A C}{B C} = \frac{1 3 x}{5 x} = \frac{1 3}{5}

Question 6

∠ A

and

∠ B

are acute angles such that

cos A = cos B

then prove that

∠ A = ∠ B

Medium

Solution

Verified by Toppr

According to the question:-

Let

cos A = \frac{s i d e a d j a c e n t A}{H y p o t e n u s e}

\frac{A C}{A B}

Similarly,

cos B = \frac{s i d e a d j a c e n t B}{H y p o t e n u s e}

= \frac{B C}{A B}

Given that

cos A = c o s B

\frac{A C}{A B} = \frac{B C}{A B}

A C = B C

In triangle,

angles opposite equal side are equal

∠ B = ∠ A

Question 7

cot θ = \frac{7}{8},

evaluate :

(i)

\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )}

(ii)

cot^{2} θ

Medium

Solution

Verified by Toppr

Given,

c o t θ = \frac{7}{8}

$tan θ = \frac{1}{cot θ} = \frac{8}{7}$

We know that,

$tan θ = \frac{o p p o s i t e S i d e}{a d j a c e n t S i d e}$

From Pythagoras theorem,

$H y p o t e n u s e^{2} = O p p o s i t e S i d e^{2} + A d j a c e n t S i d e^{2}$

$H y p o t e n u s e^{2} = 8^{2} + 7^{2}$

$H y p o t e n u s e^{2} = 64 + 49 = 113$

$H y p o t e n u s e = 113$

$sin θ = \frac{o p p o s i t e S i d e}{H y p o t e n u s e} = \frac{8}{1 1 3}$

$cos θ = \frac{A d j a c e n t S i d e}{H y p o t e n u s e} = \frac{7}{1 1 3}$

Solution(i):

$\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )}$

We have, $a^{2} - b^{2} = (a + b) (a - b)$

Similarly,

$(1 - sin^{2} θ) = (1 + sin θ) (1 - sin θ)$

$(1 - cos^{2} θ) = (1 + cos θ) (1 - cos θ)$

Therefore,

$\frac{( 1 + sin θ ) ( 1 - sin θ )}{( 1 + cos θ ) ( 1 - cos θ )} = \frac{( 1 - sin ^{2} θ )}{( 1 - cos ^{2} θ )}$

$= \frac{( 1 - ( \frac{8}{1 1 3} ) ^{2} )}{( 1 - ( \frac{7}{1 1 3} ) ^{2} )}$

$= \frac{( 1 1 3 - 6 4 )}{( 1 1 3 - 4 9 )} = \frac{4 9}{6 4}$

Solution(ii):

Given,

$c o t θ = \frac{7}{8}$

$c o t^{2} θ = (\frac{7}{8})^{2} = \frac{4 9}{6 4}$

Question 8

3 c o t A = 4

, check whether

\frac{1 - tan ^{2} A}{1 + tan ^{2} A} = cos^{2} A - sin^{2} A

or not

Yes

Can't say

Insufficient data

Medium

Solution

Verified by Toppr

Given,

3 c o t A = 4

s i n A = \frac{B C}{A C}

c o s A = \frac{A B}{A C}

t a n A = \frac{B C}{A B}

c o t A = \frac{A B}{B C} = \frac{4 k}{3 k}

A B = 4 k

B C = 3 k

By pythagoras Theorem we get,

A C^{2} = A B^{2} + B C^{2}

A C^{2} = 4^{2} + 3^{2}

A C^{2} = 16 + 9

A C^{2} = 25

A C^{2} = 25

A C = 5 k

L.H.S

= \frac{1 - t a n ^{2} A}{1 + t a n ^{2} A}

= \frac{1 - ( \frac{3}{4} ) ^{2}}{1 + ( \frac{3}{4} ) ^{2}}

= \frac{1 - \frac{9}{1 6}}{1 + \frac{9}{1 6}}

= \frac{1 6 - 9}{1 6 + 9}

= \frac{7}{2 5}

R.H.S

= cos^{2} A - sin^{2} A

= (\frac{4}{5})^{2} - (\frac{3}{5})^{2}

= \frac{1 6}{2 5} - \frac{9}{2 5}

= \frac{7}{2 5}

Hence,
L.H.S=R.H.S proved.

Question 9

In triangle

A B C

, right-angled at

B

, if

tan A = \frac{1}{3}

, find the value of:
(i)

sin A cos C + cos A sin C

(ii)

cos A cos C - sin A sin C

Medium

Solution

Verified by Toppr

△ A B C

∠ B = 9 0^{o}

tan A = \frac{1}{3} = \frac{B C}{A B}

Let

B C

= 1 x, A B = 3 x

A C^{2} = A B^{2} + B C^{2}

A C^{2} = (3 x)^{2} + (x)^{2} = 4 x^{2}

A C = 2 x

(i)

sin A cos C + cos A sin C = \frac{1}{2} \times \frac{1}{2} + \frac{3}{2} \times \frac{3}{2} = \frac{1}{4} + \frac{3}{4} = 1

(ii)

cos A cos C - sin A sin C = \frac{3}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} - \frac{3}{4} = 0

Question 10

Δ P Q R

, right-angled at Q.

P R + Q R = 25

cm and PQ=5 cm. Determine the values of

sin P, cos P, tan P

Hard

Solution

Verified by Toppr

\Rightarrow P Q = 5 c m

\Rightarrow P R + Q R = 25 c m

\Rightarrow P R = 25 - Q R

Now, In

△ P Q R

\Rightarrow (P R)^{2} = P Q^{2} + Q R^{2}

\Rightarrow (25 - Q R)^{2} = 5^{2} + Q R^{2}

\Rightarrow 625 + Q R^{2} - 50 Q R = 25 + Q R^{2}

\Rightarrow 50 Q R = 600

\Rightarrow Q R = 12 c m

\Rightarrow P R = 25 - 12 = 13 c m

∴ sin P = \frac{Q R}{P R} = \frac{1 2}{1 3}, cos P = \frac{P Q}{P R} = \frac{5}{1 3}, tan P = \frac{Q R}{P Q} = \frac{1 2}{5}

Hence, the answers are

sin P = \frac{1 2}{1 3}, cos P = \frac{5}{1 3}, tan P = \frac{1 2}{5} .

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NCERT Solutions for Class 10 Maths Chapter 8 : Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry – Brief Overview

1. Trigonometry

Trigonometry is the branch of mathematics that investigates how to use unique procedures to find distances and heights, sides, and angles within a right-angled triangle. It is a method for determining the missing angles and sides of a triangle. The term 'trigono' signifies triangle, while the word 'metry' means to measure.

2. Trigonometric Ratios

So, according to triangle characteristics, angle A represents the acute angle here. Now the side BC, opposite to the angle A is referred to as the side opposite to Angle A. In this case, AB represents the hypotenuse of a right triangle. And the side AC is referred to as the angle A's adjacent side.

The trigonometric ratios of the angle A in right triangle ABC are:

sine of ∠A = Opposite / Hypotenuse = BC / AB

cosine of ∠A = Adjacent / Hypotenuse = AC / AB

tangent of ∠A = Opposite / Adjacent = BC / AC

cosecant of ∠A = 1 / sine of ∠A = Hypotenuse / Opposite = AB / BC

secant of ∠A = 1 / cosine of ∠A = Hypotenuse / Adjacent = AB / AC

cotangent of ∠A = 1 / tangent of ∠A = Adjacent / Opposite = AC / BC

Trigonometric Ratios of Some Specific Angles

3. Trigonometric Ratios of Complementary Angles

If the total of two angles equals 90°, they are said to be complimentary. As a result, A and C are complementary angles, with ∠A + ∠C = 90°

sin (90° – A) = cos A cos (90° – A) = sin A tan (90° – A) = cot A cot (90° – A) = tan A sec (90° – A) = cosec A cosec (90° – A) = sec A

4. Trigonometric Identities

A trigonometric identity is an equation containing trigonometric ratios of an angle that is true for all values of the angle(s) involved.

sin²θ + cos²θ = 1 [for 0° ≤ θ ≤ 90°]
sec²θ – tan²θ = 1 [for 0° ≤ θ ≤ 90°]
cosec²θ – cot²θ = 1 [for 0° < θ ≤ 90°]

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles

Chapter 13 : Surface Areas and Volumes

Chapter 14 : Statistics

Chapter 15 : Probability

Frequently Asked Questions on NCERT Class 10 Maths Chapter 8 : Introduction to Trigonometry

Q1. What are the most important topics in NCERT Class 10 Chapter 8 Introduction to Trigonometry?

Answer: Chapter 8 of Class 10 Maths is mostly concerned with trigonometry, which is a critical topic in Class 10. It introduces ratios and identities, as well as trigonometric ratios of some specific angles, ratios of some complementary angles, and trigonometric identities for solving equations. This chapter must be completed methodically and with great focus by students.

Q2. How many exercises are there in Chapter 8 of Class 10 Maths?

Answer: Chapter 8 of Class 10 Maths Introduction to Trigonometry has 27 questions, 10 of which are basic, 10 of which are moderate, and 7 of which are extended answer type problems. These questions are divided into four exercises.

Q3. What is the importance of Trigonometry?

Answer: The most successful and progressive use of trigonometry is the analysis and simplification of equations utilizing various trigonometric functions such as sine, cosine, tangent, and so on. The analytical application of trigonometry is critical in engineering domains such as mechanical engineering, electronics, and mechatronics. Trigonometry can also be used to roof a house, make the roof inclined (in the case of single-family bungalows), and determine the height of the roof in buildings, among other things. It is employed in the marine and aviation sectors. It is employed in cartography (creation of maps). Trigonometry is used in satellite systems as well.

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles