**NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry**

NCERT Chapter 8 explains the relationship between the angles and sides of a right triangle. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry are required for students taking the Class 10 Board exams. These NCERT solutions for Class 10 Maths chapter 8 are strictly based on the NCERT books for Class 10 Maths. NCERT Class 10 maths solutions chapter 8 provides a thorough explanation of each question in the textbook. Furthermore, NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry provides many tips and tricks for answering questions easily. You can also look through the NCERT solutions for Class 10 for other subjects.

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

**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 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

**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.