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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry explains the relation between the angles and sides of a right angle triangle. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry are a valuable resource for students as they assist in both understanding the concepts and performing well on the CBSE Class 10 board examination. Introduction to trigonometry class 10 solutions are created by subject experts and include answers for all questions in the textbook. They are also updated to align with the latest CBSE Syllabus for 2022-23 and exam pattern.
In addition to providing a strong foundation for the concepts in this chapter, the NCERT Books class 10 trigonometry solutions also allow students to clear their doubts and grasp the fundamentals. They also provide helpful guidance for solving challenging problems in each exercise of Chapter 8 Introduction to Trigonometry.
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Also Read,
Arc Length in a Circle:
If an arc of length 'l' in a circle with radius 'r' subtends an angle of 'θ' radians.
l = rθ
θ = l/r
Conversion Between Radian and Degree Measures:
Radian Measure = (π/180) × Degree Measure
Degree Measure = (180/π) × Radian Measure
Trigonometric Ratios for Right Triangles:
In a right triangle with an angle 'θ':
sin θ = Opposite/Hypotenuse
cos θ = Adjacent/Hypotenuse
tan θ = Opposite/Adjacent
cosec θ = Hypotenuse/Opposite
sec θ = Hypotenuse/Adjacent
cot θ = Adjacent/Opposite
Reciprocal Trigonometric Ratios:
sin θ = 1/(cosec θ)
cosec θ = 1/(sin θ)
cos θ = 1/(sec θ)
sec θ = 1/(cos θ)
tan θ = 1/(cot θ)
cot θ = 1/(tan θ)
Trigonometric Ratios of Complementary Angles:
For an angle 'θ', the trigonometric ratios of its complementary angle (90° – θ) are:
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ
Trigonometric Identities:
sin^{2} θ + cos^{2} θ = 1
sin^{2} θ = 1 – cos^{2} θ
cos^{2} θ = 1 – sin^{2} θ
cosec^{2} θ – cot^{2} θ = 1
cot^{2} θ = cosec^{2} θ – 1
sec^{2} θ – tan^{2} θ = 1
tan^{2} θ = sec^{2} θ – 1
Free download NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry PDF for CBSE Exam.
Trigonometry chapter class 10 ncert solutions Excercise: 8.1
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 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
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
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 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 .
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 right-angled triangle ABC in which
Now, By using Pythagoras theorem,
So,
and
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:
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
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) [ ]
Trigonometry chapter class 10 ncert solutions Excercise: 8.2
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)
Trigonometry chapter class 10 ncert solutions Excercise: 8.3
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;
Introduction to trigonometry class 10 solutions Excercise: 8.4
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 sec^{2}A-tan^{2}A=1
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.
Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. , using the identity
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.
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.
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 |
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | Introduction to Trigonometry |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
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.
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.
Trigonometry is a most important field in mathematics which is useful in almost every field including architecture, electronics, seismology, meteorology, oceanography etc. Trigonometry problems can be solved using NCERT book and NCERT exemplar for Class 10 Mathematics. Students can download trigonometry class 10 NCERT solutions pdf for ease and study both online and offline mode.
There are a total of 15 chapters in the Class 10 Maths NCERT syllabus. Questions from all the exercises of each chapter are available in the Careers 360 website. The chapter wise link provided navigate you the solution page of the respective chapter. Students can download NCERT solutions for class 10 maths chapter 8 pdf using the link give above in this article.
The topics that are commonly covered in CBSE Maths exams for introduction to trigonometry class 10 maths chapter 8 solutions include: Introduction to Trigonometry, Trigonometric Identities, Trigonometric Ratios of Specific Angles, Trigonometric Ratios of Complementary Angles, and Trigonometric Identities. CBSE board class 10 paper is entirely based on the NCERT.
Careers360 offers precise answers to the questions found in the NCERT Solutions for class 10 chapter 8 maths. These solutions can be accessed online and also downloaded in PDF format. The solutions for maths chapter 8 class 10 are explained by experts in a clear and concise manner, and diagrams are included as needed.
Yes, All chapters in the introduction to trigonometry class 10 solutions are essential for both board exams and future grades. It's crucial for students to practise all the questions in NCERT Solutions for Class 10 Maths Chapter 8 in order to achieve high marks on the exams.
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Thanking you
Hello aspirant,
The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.
To get the complete datesheet, you can visit our website by clicking on the link given below.
https://school.careers360.com/boards/cbse/cbse-date-sheet
Thank you
Hope this information helps you.
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Thank you
Hope it helps you.
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