NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 8 Introduction to Trigonometry Class 10 NCERT Solutions Ex 8.1.

Here you get the CBSE Class 10 Mathematics chapter 8, Introduction to Trigonometry and its Applications: We are provides free comprehensive chapter wise class 10 Mathematics notes with proper images & diagram. Here you will find all the answers to the NCERT textbook questions of Chapter 8 Introduction to Trigonometry.

• Introduction to Trigonometry Class 10 Ex 8.2
• Introduction to Trigonometry Class 10 Ex 8.3
• Introduction to Trigonometry Class 10 Ex 8.4

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

NCERT Solutions for Class 10 Maths

Page No: 181

Question 1.
In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Solution:
In ∆ABC by applying Pythagoras theorem
AC² = AB² + BC²
= (24)² + (7)²
= 576 + 49
= 625
AC = √625 = 25 cm

Question 2. In Figure, find tan P – cot R.

Solution:
In  ∆PQR by applying Pythagoras theorem
PR² = PQ² + QR²
(13)² = (12)² + QR²
169 = 144 + QR²
25 = QR²
QR = 5

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

Solution:
Let  ∆ABC be a right angled triangle, right angled at point B.
Given that

sin A = 3/4
BC/AC = 3/4
Let BC be 3 K so AC will be 4 K where K ¡s a positive integer.
Now applying Pythagoras theorem in ∆ABC
AC² = AB² + BC²
(4 K)² = AB² + (3 K)²
16 K² – g K² = AB²
7 K² = AB²
AB = √7k

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

Solution:
Consider a right triangle, right angled at B

Let AB be 8 K so BC will be 15 K where K is a positive integer.
Now applying Pythagoras theorem in ∆ABC
AC² = AB² + BC²
= (8K)² + (15K)²
= 64 K² + 225 K²
= 289 K²
AC = 17 K

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

Solution:
Consider a right angle triangle ∆ABC right angled at point B.

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

Solution:
Since cos A = cos B
So AB/AC = BC/AC
AB = BC
So ∠A = ∠B (Angles opposite to equal sides are equal in length)

Question 7. If cot θ =7/8, evaluate :
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right) } \)
(ii) Cot²θ

Solution:
Consider a right angle triangle ∆ABC right angled at point B.

Question 8. If 3cot A = 4/3 , check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A } ={ cos }^{ 2 }A-{ sin }^{ 2 }A\) or not.

Solution:

Question 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

Solution:

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

Solution:

Question 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.

Solution:

Let AC be 12 K, AB will be 5 K, where K is a positive integer
Now applying Pythagoras theorem in ∆ABC
AC² = AB² + BC²
(12 K)² = (5 K)² + BC²
144 K² = 25 K² + BC²
BC² = 119 K²
BC = 109 K
We may observe that for given two sides AC = 12 K and AB = S K
BC should be such that –
AC – AB < BC < AC + AB
12 K – 5 K < BC <12 K + 5 K
7 K < BC < 17 K
But BC = 10.9 K.
Clearly such a triangle is possible and hence such value of sec A is possible. Hence, the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.

(iv) cot A is not the product of cot and A but it is cotangent of ∠A. Hence, the given statement is false.

(v) sine θ = 4/3
We know that in a right angle triangle

In a right angle triangle hypotenuse is always greater then the remaining two sides.
Hence such value of sin θ is not possible. Hence, the given statement is false.

We hope the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1, drop a comment below and we will get back to you at the earliest.

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 Triangles Class 10 NCERT Solutions Ex 6.1. 

Here you get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination as well as other competitive exams. Here you will find all the answers to the NCERT textbook questions of Chapter 6 – Triangles.

• Triangles Class 10 Ex 6.2
• Triangles Class 10 Ex 6.3
• Triangles Class 10 Ex 6.4
• Triangles Class 10 Ex 6.5
• Triangles Class 10 Ex 6.6

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

NCERT Solutions for Class 10 Maths

Page No: 122

Question 1.
Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are __________ and
(b) their corresponding sides are __________. (equal, proportional)

Solution:
(i) All circles are SIMILAR.
(ii) All squares are SIMILAR.
(iii) All EQUILATERAL triangles are similar.
(iv) Two polygons of same number of sides are similar, if their corresponding angles are EQUAL and their corresponding sides are PROPORTIONAL.

Question 2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Solution:
(i)  Two equilateral triangles with sides 1 cm and 2 cm.

Two squares with sides 1 cm and 2 cm

(ii)  Trapezium and Square

Triangle and Parallelogram

Question 3.
State whether the following quadrilaterals are similar or not:

Solution:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. 1:2 but their corresponding angles are not equal.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 10 Circles Class 10 NCERT Solutions Ex 10.1.

Class 10th Math’s important questions in chapter Area Related to Circles for free download in PDF format. The most important questions for annual examination from chapter Introduction to Area Related to Circles are given here for download. Here you will find all the answers to the NCERT textbook questions of Chapter 10 Circles.

• Circles Class 10 Ex 10.2

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1

NCERT Solutions for Class 10 Maths

Page No: 209

Question 1. How many tangents can a circle have?

Solution:
A circle can have infinite tangents.

Question 2. Fill in the blanks :
(i) A tangent to a circle intersects it in …………… point(s).
(ii) A line intersecting a circle in two points is called a ………….
(iii) A circle can have …………… parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called …………

Solution:
(i) one
(ii) secant
(iii) two
(iv) point of contact

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm

Solution:

PQ = √119 cm
Hence, (D) is the correct answer.

Concept Insight: To answer such type of problems, remember to use the result that the radius is perpendicular to the tangent at the point of contact and then make use of pythagoras theorem in the right traingle.

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution:

AB and CD are two lines parallel to the line XY. Line AB is intersecting the circle at exactly two points, P and Q. So, line AB is secant of this circle. The line CD is intersecting the circle at exactly one point, R. So, line CD is tangent to circle.

Concept Insight: To draw the two lines, remember that a tangent touches the circle at one point only and a secant intersects the circle at two distinct points.

We hope the NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions For Class 10 Maths – AplusTopperhttps://t.co/q7hSKh3xmf#NCERTSolutionsForClass10Maths #NCERTSolutions #AplusTopper

— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 2 Polynomials Class 10 NCERT Solutions Ex 2.1. 

NCERT Solutions for Class 10 maths chapter 2 Polynomials is available here. It includes solutions to all the questions present in class 10 Maths NCERT textbooks. These solutions are prepared by our subject experts to help students in their academics. All questions and answers from the NCERT Book of class lo Math Chapter 2 are provided here for you for free.

• Polynomials Class 10 Ex 2.2
• Polynomials Class 10 Ex 2.3
• Polynomials Class 10 Ex 2.4

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1

Page No: 28

Question 1.
The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solution:
(i) The graph of P(x) does not cut the x-axis at all.
So, the number of zeroes is 0.
(ii) The graph of P(x) intersects the x-axis at only 1 point.
So, the number of zeroes is 1.
(iii) The graph of P(x) intersects the x-axis at 3 points.
So, the number of zeroes is 3.
(iv)  The graph of P(x) intersects the x-axis at 2 points.
So, the number of zeroes is 2.
(v) The graph of P(x)  intersects the x-axis at 4 points.
So, the number of zeroes is 4.
(vi) The graph of P(x) intersects the x-axis at 3 points.
So, the number of zeroes is 3.

Concept Insight: Since the polynomial p(x) given here is a polynomial in variable x, so to find the number of zeroes, we look at the number of points where the graph intersects or touches the x-axis and not the y-axis.
At all these points where the graph intersects x axis the value of the polynomial y = p(x) will be zero.

CBSE Class 10 Previous Year Question Papers – Download PDF

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 are part of
NCERT Solutions for Class 10 Maths. Here are we have given Chapter 5 Arithmetic Progressions Class 10 NCERT Solutions Ex 5.1. 

Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination as well as other competitive exams.

• Arithmetic Progressions Class 10 Ex 5.2
• Arithmetic Progressions Class 10 Ex 5.3
• Arithmetic Progressions Class 10 Ex 5.4

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Page No: 99

Question 1.
In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution :

Concept Insight: To crack such problems read the question carefully, analyse and then write the data the initial value changes in a particular pattern. But the sequence represents an arithmetic progression if the difference of any two consecutive terms is constant.

Question 2.
Write first four terms of the A.P. when the first term a and the common difference d are given as follows
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Solution :

Concept Insight: Remember the basic definition of an AP it can be generated given its first term and common difference by  adding the common difference to the previous term i.e a,a+d,a+2d,….a+(n-1)d.

Question 3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …

Solution :

Concept Insight: Remember the basic definition of an AP common difference is the difference of consecutive terms, other terms can be generated by following general pattern a,a+d,a+2d,…. a+(n-1)d

Question 4.
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴ …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7² …
(xv) 1², 5², 7², 7³ …

Solution :