NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry - Chapter 8 of NCERT explains the relation between the angles and sides of a right angle triangle. Students appearing in the Class 10 Board exams must check the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry. These Class 10 Maths chapter 8 NCERT solutions are strictly based on the NCERT books for Class 10 Maths. NCERT Class 10 maths solutions chapter 8 gives a detailed explanation of each and every question given in the textbook. Moreover, NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry offers many tips and tricks to solve the questions in an easy way. You can also check the NCERT solutions for Class 10 for other subjects as well.

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  • Trigonometry Class 10 NCERT Exemplar Solutions
  • Notes For Trigonometry Class 10

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In , right-angled at , . Determine :

Answer:

We have,
In , B = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm
So, by using Pythagoras theorem,

Therefore,

AC = 25 cm

Now,
(i)

(ii) For angle C, AB is perpendicular to the base (BC). Here B indicates to Base and P means perpendicular wrt angle C
So,
and

Q2 In Fig. 8.13, find .

Answer:


We have, PQR is a right-angled triangle, length of PQ and PR are 12 cm and 13 cm respectively.
So, by using Pythagoras theorem,


Now, According to question,
=
= 5/12 - 5/12 = 0

Q3 If calculate and .

Answer:

Suppose ABC is a right-angled triangle in which and we have
So,

Let the length of AB be 4 unit and the length of BC = 3 unit So, by using Pythagoras theorem,
units
Therefore,
and

Q4 Given find and .

Answer:

We have,

It implies that In the triangle ABC in which . The length of AB be 8 units and the length of BC = 15 units

Now, by using Pythagoras theorem,

units

So,
and

Q5 Given calculate all other trigonometric ratios.

Answer:

We have,

It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which B is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.

By using Pythagoras theorem,

BC = 5 unit

Therefore,

Q6 If and are acute angles such that , then show that .

Answer:

We have, A and B are two acute angles of triangle ABC and

According to question, In triangle ABC,




Therefore, A = B [angle opposite to equal sides are equal]

Q7 If evaluate:

Answer:

Given that,

perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which
Now, By using Pythagoras theorem,

So,
and





Q8 If check wether or not.

Answer:

Given that,


ABC is a right-angled triangle in which and the length of the base AB is 4 units and length of perpendicular is 3 units

By using Pythagoras theorem, In triangle ABC,

AC = 5 units

So,




Put the values of above trigonometric ratios, we get;


LHS RHS

Q9 In triangle , right-angled at , if find the value of:


Answer:

Given a triangle ABC, right-angled at B and

According to question,
By using Pythagoras theorem,

AC = 2
Now,

Therefore,




Q10 In , right-angled at , and . Determine the values of

Answer:

We have, PR + QR = 25 cm.............(i)
PQ = 5 cm
and
According to question,
In triangle PQR,
By using Pythagoras theorem,

PR - QR = 1........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm and QR = 12 cm.

therefore,

Q11 State whether the following are true or false. Justify your answer.

(i) The value of is always less than 1.
(ii) for some value of angle A.
(iii) is the abbreviation used for the cosecant of angle A.
(iv) is the product of cot and A.
(v) for some angle

Answer:

(i) False,
because , which is greater than 1

(ii) TRue,
because

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

(iv) False,
because the term is a single term, not a product.

(v) False,
because lies between (-1 to +1) [ ]


NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.2

Q1 Evaluate the following :

Answer:


As we know,
the value of ,


Q1 Evaluate the following :

Answer:

We know the value of

and


According to question,


Q1 Evaluate the following :

Answer:


we know the value of

, and ,

After putting these values



Q1 Evaluate the following :

Answer:

..................(i)
It is known that the values of the given trigonometric functions,

Put all these values in equation (i), we get;

Q1 Evaluate the following :

Answer:

.....................(i)
We know the values of-

By substituting all these values in equation(i), we get;

Q2 Choose the correct option and justify your choice :

Answer:

Put the value of tan 30 in the given question-

The correct option is (A)

Q2 Choose the correct option and justify your choice :

Answer:

The correct option is (D)

We know that
So,

Q2 Choose the correct option and justify your choice :

is true when =

Answer:

The correct option is (A)

We know that
So,

Q2 Choose the correct option and justify your choice :

Answer:

Put the value of

The correct option is (C)

Q3 If and find

Answer:

Given that,

So, ..........(i)

therefore, .......(ii)
By solving the equation (i) and (ii) we get;

and

Q4 State whether the following are true or false. Justify your answer.


The value of increases as increases.
The value of increases as increases.
for all values of .
is not defined for

Answer:

(i) False,
Let A = B =
Then,


(ii) True,
Take
whent
= 0 then zero(0),
= 30 then value of is 1/2 = 0.5
= 45 then value of is 0.707


(iii) False,


(iv) False,
Let = 0


(v) True,
(not defined)


NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Q1 Evaluate :

Answer:


We can write the above equation as;

By using the identity of
Therefore,

So, the answer is 1.

Q1 Evaluate :

Answer:


The above equation can be written as ;

.........(i)
It is known that,
Therefore, equation (i) becomes,

So, the answer is 1.

Q1 Evaluate :

Answer:


The above equation can be written as ;
....................(i)
It is known that
Therefore, equation (i) becomes,

So, the answer is 0.

Q1 Evaluate :

Answer:

This equation can be written as;
.................(i)
We know that

Therefore, equation (i) becomes;
= 0

So, the answer is 0.

Q2 Show that :

Answer:


Taking Left Hand Side (LHS)
=

[it is known that and

Hence proved.

Q2 Show that :

Answer:

Taking Left Hand Side (LHS)
=
=
= [it is known that and ]
= 0

Q3 If , where is an acute angle, find the value of .

Answer:

We have,
2A = (A - )
we know that,

Q4 If , prove that .

Answer:

We have,

and we know that
therefore,

A = 90 - B
A + B = 90
Hence proved.

Q5 If , where is an acute angle, find the value of .

Answer:

We have,
, Here 4A is an acute angle
According to question,
We know that

Q6 If and are interior angles of a triangle , then show that

Answer:

Given that,
A, B and C are interior angles of
To prove -

Now,
In triangle ,
A + B + C =




Hence proved.

Q7 Express in terms of trigonometric ratios of angles between and .

Answer:

By using the identity of and

We know that,
and
the above equation can be written as;


NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

Q1 Express the trigonometric ratios and in terms of .

Answer:

We know that
(i)

(ii) We know the identity of

(iii)

Q2 Write all the other trigonometric ratios of in terms of .

Answer:

We know that the identity



Q3 Evaluate :

Answer:

....................(i)

The above equation can be written as;


(Since )

Q3 Evaluate :

Answer:

We know that

Therefore,

Q4 Choose the correct option. Justify your choice.

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

.............(i)

and it is known that sec2A-tan2A=1

Therefore, equation (i) becomes,

Q4 Choose the correct option. Justify your choice.

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

.......................(i)

we can write his above equation as;

= 2

Q4 Choose the correct option. Justify your choice.

Answer:

The correct option is (D)


Q4 Choose the correct option. Justify your choice.

Answer:

The correct option is (D)

..........................eq (i)

The above equation can be written as;

We know that

therefore,

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

We need to prove-

Now, taking LHS,



LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

We need to prove-

taking LHS;

= RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

[ Hint: Write the expression in terms of and ]

Answer:

We need to prove-

Taking LHS;


By using the identity a 3 - b 3 =(a - b) (a 2 + b 2 +ab)

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

taking LHS;

Taking RHS;
We know that identity

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. , using the identity

Answer:

We need to prove -

Dividing the numerator and denominator by , we get;

Hence Proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

We need to prove -

Taking LHS;
By rationalising the denominator, we get;

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

We need to prove -

Taking LHS;
[we know the identity ]

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

Given equation,
..................(i)

Taking LHS;



[since ]

Hence proved

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

Taking LHS;

Taking RHS;

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer:

We need to prove,

Taking LHS;

Taking RHS;

LHS = RHS

Hence proved.

Features of Trigonometry Class 10 NCERT Solutions

Unit 5 "Trigonometry" holds 12 marks out of 80 marks in the maths paper of CBSE board examination and we can expect 2-3 questions from this chapter of total around 8 marks. There is a total of 4 exercises with 27 questions in the NCERT solutions for class 10 maths chapter 8. These NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry are designed to provide assistance for homework and for preparing the board examinations.

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Trigonometry Exercise-Wise Solutions

  • Trigonometry Class 10 Ex-8.1
  • Trigonometry Class 10 Ex-8.2
  • Trigonometry Class 10 Ex-8.3
  • Trigonometry Class 10 Ex-8.4

Trigonometry Class 10 Topic-

The trigonometric ratios of the angle A in right triangle ABC are defined as follows-

The values of all the trigonometric ratios of 0°, 30°, 45°, 60°, and 90° are-

Sin A

0

1

Cos A

1


0

Tan A

0

1

Not defined

Cosec A

Not defined

2

1

Sec A

1

2

Not defined

Cot A

Not defined

1

0

NCERT Solutions for Class 10 Maths All Chapters

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 chapter 5 Arithmetic Progressions

Chapter 6

NCERT solutions for class 10 maths chapter 6 Triangles

Chapter 7

NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions for class 10 maths chapter 11 Constructions

Chapter 12

NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

NCERT solutions for class 10 maths chapter 15 Probability

Benefits of NCERT Solutions for Class 10 Maths Chapter 8

  • These Class 10 Maths Chapter 8 NCERT solutions are prepared by the experts. Hence these solutions are 100 per cent reliable.

  • The Trigonometry Class 10 will be beneficial for Class 10 board exams and for higher studies as well.

  • NCERT chapter 8 Maths Class 10 solutions will help in building the basic concepts of trigonometry and bring forth some easy ways to solve the questions.

Subject-wise NCERT Solutions of Class 10

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How to use NCERT solutions for Class 10 Maths chapter 8 Introduction to Trigonometry?

  • Firstly, learn all the concepts given in the NCERT book. Memorise all the trigonometric ratios, angle values, and trigonometric identities.

  • Now practice exercises by referring to the NCERT Class 10 Maths solutions chapter 8.

  • As the NCERT Solutions for Class 10 Maths Chapter 8 PDF Download is not available. So you can save the webpage to practice the solutions offline.

  • After doing all these you can practice the last 5 years question papers of board examinations.

Subject wise NCERT Exemplar solutions

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