NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry  The chapter trigonometry has been generated from the right angle triangle. The relation between the angles and sides of a right angle triangle is called trigonometry. In solutions of NCERT class 10 maths chapter 8 Introduction to Trigonometry, there is a detailed explanation for each question of practice exercise. In this particular chapter, you will study some ratios of a rightangled triangle with respect to its acute angles, and this ratio is called a trigonometric ratio of the angles. CBSE NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry is also covering the questions based on trigonometric ratios. Unit 5 "Trigonometry" holds 12 marks out of 80 marks in the maths paper of CBSE board examination and we can expect 23 questions from this chapter of total around 8 marks. There is a total of 4 exercises with 27 questions in the entire chapter. NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry are designed to provide assistance for homework and for preparing the board examinations. If you want to download the NCERT solutions for free then you can have it by clicking on the link.
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 
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
Answer:
We have, PQR is a rightangled 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
Suppose ABC is a rightangled 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
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.
We have,
It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a rightangled 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 .
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:
Given that,
perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a rightangled triangle ABC in which
Now, By using Pythagoras theorem,
So,
and
Given that,
ABC is a rightangled 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 , rightangled at , if find the value of:
Given a triangle ABC, rightangled at B and
According to question,
By using Pythagoras theorem,
AC = 2
Now,
Therefore,
Q10 In , rightangled at , and . Determine the values of
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
(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) []
As we know,
the value of ,
..................(i)
It is known that the values of the given trigonometric functions,
Put all these values in equation (i), we get;
.....................(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 :
Put the value of tan 30 in the given question
The correct option is (A)
Q2 Choose the correct option and justify your choice :
The correct option is (D)
We know that
So,
Q2 Choose the correct option and justify your choice :
is true when =
The correct option is (A)
We know that
So,
Q2 Choose the correct option and justify your choice :
Put the value of
The correct option is (C)
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
(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)
Q1 Evaluate :
We can write the above equation as;
By using the identity of
Therefore,
So, the answer is 1.
Q1 Evaluate :
The above equation can be written as ;
.........(i)
It is known that,
Therefore, equation (i) becomes,
So, the answer is 1.
Q1 Evaluate :
The above equation can be written as ;
....................(i)
It is known that
Therefore, equation (i) becomes,
So, the answer is 0.
Q1 Evaluate :
This equation can be written as;
.................(i)
We know that
Therefore, equation (i) becomes;
= 0
So, the answer is 0.
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 .
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
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 .
By using the identity of and
We know that,
and
the above equation can be written as;
Q1 Express the trigonometric ratios and in terms of .
We know that
(i)
(ii) We know the identity of
(iii)
Q4 Choose the correct option. Justify your choice.
(A) 1 (B) 9 (C) 8 (D) 0
The correct option is (B) = 9
.............(i)
and it is known that
Therefore, equation (i) becomes,
Q4 Choose the correct option. Justify your choice.
(A) 0 (B) 1 (C) 2 (D) –1
The correct option is (C)
.......................(i)
we can write his above equation as;
= 2
Q4 Choose the correct option. Justify your choice.
The correct option is (D)
..........................eq (i)
The above equation can be written as;
We know that
therefore,
We need to prove
Now, taking LHS,
LHS = RHS
Hence proved.
We need to prove
taking LHS;
= RHS
Hence proved.
[Hint : Write the expression in terms of and ]
We need to prove
Taking LHS;
By using the identity a^{3}  b^{3} =(a  b) (a^{2} + b^{2}+ab)
Hence proved.
[Hint : Simplify LHS and RHS separately]
We need to prove
taking LHS;
Taking RHS;
We know that identity
LHS = RHS
Hence proved.
We need to prove 
Dividing the numerator and denominator by , we get;
Hence Proved.
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.
We need to prove 
Taking LHS;
[we know the identity ]
Hence proved.
Given equation,
..................(i)
Taking LHS;
[since ]
Hence proved
[Hint : Simplify LHS and RHS separately]
We need to prove
Taking LHS;
Taking RHS;
LHS = RHS
Hence proved.
We need to prove,
Taking LHS;
Taking RHS;
LHS = RHS
Hence proved.
Chapter No. 
Chapter Name 
Chapter 1 
CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers 
Chapter 2 

Chapter 3 
Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4 
CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5 
NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6 

Chapter 7 
CBSE 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 
Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10 

Chapter 11 

Chapter 12 
Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13 
CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14 

Chapter 15 
First of all, be familiar with the concepts of the rightangle triangle.
This chapter needs high memorization means you have to memorize the trigonometric ratios, angle values, and trigonometric identities.
Once you have done the basic part then you can jump to the practice exercises.
While doing the practice exercises if you stuck anywhere then you take the help of NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry.
After doing all these you can practice the last 5 years question papers of board examinations.
Keep working hard & happy learning!