NCERT Solutions for Class 10 Maths Exercise 8.1 Chapter 8 Introduction to Trigonometry – FREE PDF Download

NCERT Class 10 Maths Ch 8 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 8 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Introduction to Trigonometry solutions will help you understand the chapter thoroughly.

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

1. In ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i)

(ii)

Ans. Let us draw a right angled triangle ABC, right angled at B.

Using Pythagoras theorem,

Let AC = 24k and BC = 7k

Using Pythagoras theorem,

= = 576 + 49 = 625

AC = 25 cm

(i) $\sin A = \frac{P}{H} = \frac{B C}{A C} = \frac{7}{25}$ , $\cos A = \frac{B}{H} = \frac{A B}{A C} = \frac{24}{25}$

(ii) $\sin C = \frac{P}{H} = \frac{A B}{A C} = \frac{24}{25}$ , $\cos C = \frac{B}{H} = \frac{B C}{A C} = \frac{7}{25}$

2. In adjoining figure, find :

Ans. In triangle PQR, Using Pythagoras theorem,

= 169 – 144 = 25

QR = 5 cm

= $\frac{P}{B} - \frac{B}{P}$ = = = 0

3. If calculate and

Ans. Given: A triangle ABC in which

B =

Let BC = and AC =

Then, Using Pythagoras theorem,

AB = =

= =

$\cos A = \frac{B}{H} = \frac{A B}{A C} = \frac{k \sqrt{7}}{4 k} = \frac{\sqrt{7}}{4}$

$\tan A = \frac{P}{B} = \frac{B C}{A B} = \frac{3 k}{k \sqrt{7}} = \frac{3}{\sqrt{7}}$

4. Given find and

Ans. Given: A triangle ABC in which

B =

Let AB = and BC =

Then using Pythagoras theorem,

AC =

= =

$\sin A = \frac{P}{H} = \frac{B C}{A C} = \frac{15 k}{17 k} = \frac{15}{17}$

$\sec A = \frac{H}{B} = \frac{A C}{A B} = \frac{17 k}{8 k} = \frac{17}{8}$

5. Given calculate all other trigonometric ratios.

Ans. Consider a triangle ABC in which

A =

and

B =

Let AB = and BC =

Then, using Pythagoras theorem,

BC =

= =

$\sin θ = \frac{P}{H} = \frac{B C}{A C} = \frac{5 k}{13 k} = \frac{5}{13}$

$\cos θ = \frac{B}{H} = \frac{A B}{A C} = \frac{12 k}{13 k} = \frac{12}{13}$

$\tan θ = \frac{P}{B} = \frac{B C}{A B} = \frac{5 k}{12 k} = \frac{5}{12}$

$\cot θ = \frac{B}{P} = \frac{A B}{B C} = \frac{12 k}{5 k} = \frac{12}{5}$

$\cos e c θ = \frac{H}{P} = \frac{A C}{B C} = \frac{13 k}{5 k} = \frac{13}{5}$

6. If And B are acute angles such that then show that A = B.

Ans. In right triangle ABC,

and

But [Given]

AC = BC

A = B

[Angles opposite to equal sides are equal]

7. If evaluate:

(i)

(ii)

Ans. Consider a triangle ABC in which

A =

and

B =

Let AB = and BC =

Then, using Pythagoras theorem,

AC =

= =

(i) =

= = =

(ii) = =

8. If check whether or not.

Ans. Consider a triangle ABC in which

B =

And

Let AB = and BC =

Then, using Pythagoras theorem,

AC =

= =

And

Now, L.H.S. =

= =

R.H.S. =

= =

L.H.S. = R.H.S.

9. In ABC right angles at B, if find value of:

(i)

(ii)

Ans. Consider a triangle ABC in which

B =

Let BC = and AB =

Then, using Pythagoras theorem,

AC =

= = =

For C, Base = BC, Perpendicular = AB and Hypotenuse = AC

$\cos C = \frac{B C}{A C} = \frac{k}{2 k} = \frac{1}{2}$

(i) =

= = 1

(ii) =

10. In PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of and

Ans. In

PQR, right angled at Q.

PR + QR = 25 cm and PQ = 5 cm

Let QR = cm, then PR = cm

Using Pythagoras theorem,

RQ = 12 cm and RP = 25 – 12 = 13 cm

And

11. 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 and A.

(v) for some angle

Ans. (i) False because sides of a right triangle may have any length, so

may have any value.

(ii) True as is always greater than 1.

(iii) False as is the abbreviation of cosine A.

(iv) False as is not the product of ‘cot’ and A. ‘cot’ is separated from A has no meaning.

(v) False as cannot be > 1.

NCERT Solutions class 10 Maths Exercise 8.1 Ch 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Exercise 8.1 Chapter 8 Introduction to Trigonometry – FREE PDF Download

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