CBSE Class 10 – Page 5 – MCQ Questions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-1-ex-1-3/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Real Numbers
Exercise	Ex 1.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Exercise 1.3 Class 10 Maths Question 1.
Prove that √5 is irrational.
Solutions:
Let us assume that is rational.
∴ There exists co-prime integers a and b (b ≠ 0) such that
√5 = \(\frac { a }{ b }\) ⇒ √5b= 0
Squaring on both sides, we get
5b²= a²…… (i)
⇒ 5 divides a² ⇒ 5 divides a
So, we can write a = 5c for some integer c.
From (i) and (ii)
5b² = 25c²
⇒ b² = 5c²
⇒ 5 divides b²
⇒ 5 divides b
∴ 5 is a common factor of a and b.
But this contradicts the fact that a and b are co-primes.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
Hence, √5 is irrational.

1.3 Class 10 Maths Real Numbers Question 2.
Prove that 3 + 2√5 is irrational.
Solutions:
Let us assume that 3 + 2√5 is rational.
∴ There exists co-prime integers a and b(b ≠ 0) such that
Exercise 1.3 Class 10 Maths
But this contradicts the fact that √5 is irrational.
This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational. Hence, we conclude that 3 + 2√5 is irrational.

Exercise 1.3 Class 10 Real Numbers Question 3.
Prove that the following are irrationals.
1.3 Class 10 Maths Real Numbers
Solutions:

Exercise 1.3 Real Numbers Class 10 Maths

We hope the NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

February 1, 2022February 1, 2022 by Dattu

NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-1-ex-1-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Real Numbers
Exercise	Ex 1.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Ex 1.2 Class 10 Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429.
Solutions:
(i) 140

(ii) 156
Exercise 1.2 Class 10 Maths

(iii) 3825
1.2 Class 10 Maths Chapter 1

(iv) 5005

Exercise 1.2 Class 10
(v) 7429.
So, 7429 = 17 x 19 x 23
Ex 1.2 Class 10

Exercise 1.2 Class 10 Maths Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = product of two numbers,
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solutions:
(i) 26 and 91
Class 10 Maths Chapter 1 Exercise 1.2

(ii) 510 and 92
Ex 1.2 Class 10 Maths

(iii) 336 and 54
Ex 1.2 Class 10 Maths Exercise 1.2 Class 10 Maths

1.2 Class 10 Maths Chapter 1 Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12,15 and 21
(ii) 17,23 and 29
(iii) 8, 9 and 25
Solutions:
(i) 12,15 and 21
Exercise 1.2 Class 10 Maths

(ii) 17,23 and 29
Class 10 Ex 1.2

(iii) 8, 9 and 25
Class 10 Maths Ex 1.2

Exercise 1.2 Class 10 Question 4.
Given that HCF (306,657) = 9, find LCM (306, 657).
Solutions:
Given that HCF (306, 657) = 9
We know that LCM x HCF = Product of two numbers
Class 10 Maths Chapter 1 Exercise 1.2 Solutions

Class 10 Maths Chapter 1 Exercise 1.2 Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solutions:
Since prime factorisation of 6ⁿ is given by 6ⁿ= (2 x 3)ⁿ = 2ⁿ x 3ⁿ
Prime factorisation of 6ⁿ contains only prime numbers 2 and 3.
6ⁿmay end with the digit 0 for some ‘n’ if 5 must be in its prime factorisation which is not present.
So, there is no natural number VT for which 6ⁿ ends with the digit zero.

Ex 1.2 Class 10 Maths Question 6.
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solutions:

Both N₁ and N₂ are expressed as a product of primes. Therefore, both are composite numbers.

Exercise 1.2 Class 10 Maths Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solutions:
By taking LCM of time taken (in minutes) by Sonia and Ravi, we can get the actual number of minutes after which they meet again at the starting point after both start at same point and of the same time, and go in the same direction.
Class 10 Maths Chapter 1 Exercise 1.2 Solutions

Therefore, both Sonia and Ravi will meet again at the starting point after 36 minutes.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

January 18, 2022January 18, 2022 by Prasanthi

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-9/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 9
Chapter Name	Some Applications of Trigonometry
Exercise	Ex 9.1
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 1
Solution:
Given: length of the rope (AC) = 20 m, ∠ACB = 30°
Let height of the pole (AB) = h metres

Hence, height of the pole = 10 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let DB is a tree and AD is the broken part of it which touches the ground at C.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 4

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
Let l₁ is the length of slide for children below the age of 5 years and l₂ is the length of the slide for elder children
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 5

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Solution:
Let h be the height of the tower
In ∆ABC,
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 6

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution:
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 7

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 8

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution:
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 9

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
Let the height of the pedestal AB = h m
Given: height of the statue = 1.6 m, ∠ACB = 45° and ∠DCB = 60°
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 10

Question 9.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Given: height of the tower AB = 50 m,
∠ACB = 60°, ∠DBC = 30°
Let the height of the building CD = x m
In ∆ABC,
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 11

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Solution:
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 13

Question 11.
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 14
Solution:
Let the height of the tower AB = h m and BC be the width of the canal.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let height of the tower AB = (h + 7) m
Given: CD = 7 m (height of the building),
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 17

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Given: height of the lighthouse = 75 m
Let C and D are the positions of two ships.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 19

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 22
Solution:
Let the first position of the balloon is A and after sometime it will reach to the point D.
The vertical height ED = AB = (88.2 – 1.2) m = 87 m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let the height of the tower AB = h m
Given: ∠XAD = ∠ADB = 30°
and ∠XAC = ∠ACB = 60°
Let the speed of the car = x m/sec
Distance CD = 6 x x = 6x m
Let the time taken from C to B = t sec.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 24

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 27

We hope the NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

January 18, 2022January 18, 2022 by Dattu

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-8-ex-8-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 8
Chapter Name	Introduction to Trigonometry
Exercise	Ex 8.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Class 10 Ex 8.2 Question 1.
Evaluate the following:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 1
Solution:

Ex 8.2 Class 10 Question 2.
Choose the correct option and justify your choice:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 4
Solution:

Exercise 8.2 Class 10 Question 3.
If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \surd 3 }\); 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = √3
⇒ tan (A + B) = tan 60°
⇒ A + B = 60° ……(i)
tan (A – B) = \(\frac { 1 }{ \surd 3 }\)
⇒ tan (A – B) = tan 30°
⇒ A – B = 30° ……..(ii)
Adding equation (i) and (ii), we get
2A = 90° ⇒ A = 45°
From (i), 45° + B = 60° ⇒ B = 60° – 45° = 15°
Hence, ∠A = 45°, ∠B = 15°

Exercise 8.2 Class 10 NCERT Solution Question 4.
State whether the following statements are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 7

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

January 18, 2022January 18, 2022 by Raju

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-8/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 8
Chapter Name	Introduction to Trigonometry
Exercise	Ex 8.1
Number of Questions Solved	11
Category	NCERT Solutions

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

Exercise 8.1 Class 10 Question 1.
In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 1

Class 10 Ex 8.1 Question 2.
In given figure, find tan P – cot R.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 2

Trigonometry Class 10 Question 3.
If sin A = \(\frac { 3 }{ 4 }\) , calculate cos A and tan A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 3
sin A = \(\frac { 3 }{ 4 }\)
sin A = \(\frac { BC }{ AC }\) = sin A = \(\frac { 3 }{ 4 }\)
Let BC = 3k and AC = 4k

Ex 8.1 Class 10 Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 5

Class 10 Trigonometry Ex 8.1 Question 5.
Given sec θ = \(\frac { 13 }{ 12 }\) , calculate all other trigonometric ratios.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 6

Ex 8.1 Class 10 Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 7

Ex 8.1 Class 10 Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), 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:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 8

Trigonometry NCERT Class 10 Ex 8.1 Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 9

Ex 8.1 Class 10 Maths Solutions Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac { 1 }{ \surd 3 }\), find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 10

Class 10 Ex 8.1 Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 11

NCERT Solutions For Class 10 Maths Chapter 8 Question 11.
State whether the following statements are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac { 12 }{ 5 }\) 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 = \(\frac { 4 }{ 3 }\) for some angle.
Solution:
(i) tan 60° = √3 , Since √3 > 1. (False)
(ii) sec A is always ≥ 1. (True)
(iii) cos A is the abbreviation for cosine A. (False)
(iv) cot without ∠A is meaningless. (False)
(v) sin θ can never be greater than 1.
∴ sin θ = \(\frac { P }{ H }\) , hypotenuse is always greater than other two sides. (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.