# Introduction To Trigonometry Chapter 8 Class 10 Maths NCERT PDF

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### Chapter 8: Introduction to Trigonometry

#### 8.1 Introduction

You have already studied triangles, and in particular, right triangles, in your earlier classes. Let us take some examples from our surroundings where right triangles can be imagined to be formed

#### 8.2 Trigonometric Ratios

In Section 8.1, you have seen some right triangles imagined to be formed in different situations.

#### 8.5 Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved.

Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved. In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities

### NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

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:

In a given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.

By applying the Pythagoras theorem, we get

AC2=AB2+BC2

AC2 = (24)2+72

AC2 = (576+49)

AC2 = 625cm2

AC = √625 = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25

ii) To find Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25

2. In Fig. 8.13, find tan P – cot R

Solution:

In the given triangle QPR, the given triangle is right-angled at Q and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is a right-angled triangle, to find the side QR, apply the Pythagorean theorem

According to the Pythagorean theorem,

In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.

PR2 = QR2 + PQ2

Substitute the values of PR and PQ

13= QR2+122

169 = QR2+144

Therefore, QR= 169−144

QR= 25

QR = √25 = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) Opposite side /Adjacent side = QR/PQ = 5/12

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12

Therefore,

tan (P) – cot (R) = 5/12 – 5/12 = 0

Therefore, tan(P) – cot(R) = 0