ICSE Class 9 Maths Chapter 22 Trigonometrical Ratios

Unit 7: Trigonometry

Chapter 22: Trigonometrical Ratios

Sine, Cosine, Tangent of an Angle and Their Reciprocals

22.1 Introduction

The word "Trigonometry" means measurement of triangles.

In this unit, we shall be dealing with the relations of sides and angles of right-angled triangles only.

22.2 Concept of Perpendicular, Base and Hypotenuse in a Right Triangle

For any acute angle (which is also known as the angle of reference) in a right-angled triangle: the side opposite to the acute angle is called the perpendicular; the side adjacent to it is called the base and the side opposite to the right angle is called the hypotenuse.

Examples:

(i) Angle of reference = ∠A, Perpendicular = BC, Base = AB, Hypotenuse = AC

(ii) Angle of reference = ∠C, Perpendicular = AB, Base = BC, Hypotenuse = AC

(iii) Angle of reference = α, Perpendicular = PQ, Base = QR, Hypotenuse = PR

22.3 Notation of Angles

To indicate an angle, any letter of the English alphabet can be used, but in trigonometry, in general, the following Greek letters are used:

(i) θ (theta) (ii) φ (phi) (iii) α (alpha) (iv) β (beta) (v) γ (gamma), etc.

22.4 Trigonometrical Ratios

The ratio between the lengths of a pair of two sides of a right-angled triangle is called a trigonometrical ratio.

The three sides of a right-angled triangle give six trigonometrical ratios; namely: sine, cosine, tangent, cotangent, secant and cosecant. In short, these ratios are written as: sin, cos, tan, cot, sec and cosec respectively.

In a right-angled triangle ABC, for acute angle A:

(1) sine (sin) is defined as the ratio between the lengths of perpendicular and hypotenuse.

\[\sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC}\]

(2) cosine (cos) is defined as the ratio between the lengths of base and hypotenuse.

\[\cos A = \frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC}\]

(3) tangent (tan) is defined as the ratio between the lengths of perpendicular and base.

\[\tan A = \frac{\text{perpendicular}}{\text{base}} = \frac{BC}{AB}\]

(4) cotangent (cot) is defined as the ratio between the lengths of base and perpendicular.

\[\cot A = \frac{\text{base}}{\text{perpendicular}} = \frac{AB}{BC}\]

(5) secant (sec) is defined as the ratio between the lengths of hypotenuse and base.

\[\sec A = \frac{\text{hypotenuse}}{\text{base}} = \frac{AC}{AB}\]

(6) cosecant (cosec) is defined as the ratio between the lengths of hypotenuse and perpendicular.

\[\cosec A = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{AC}{BC}\]

Similarly, for acute angle C in the given right triangle:

\[\sin C = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{AB}{AC}\]

\[\cos C = \frac{\text{base}}{\text{hypotenuse}} = \frac{BC}{AC}\]

\[\tan C = \frac{\text{perpendicular}}{\text{base}} = \frac{AB}{BC}\]

and so on.

Each trigonometrical ratio is a real number and has no unit.

Example 1

From the given figure, find:

(i) sin A (ii) cos C (iii) tan A (iv) cosec C (v) \(\sec^2 A - \tan^2 A\)

Solution:

Given, angle ABC = 90°.

\[AC^2 = AB^2 + BC^2 \text{ (AC is hyp.)}\]

\[13^2 = 5^2 + BC^2\]

\[BC^2 = 169 - 25 = 144 \text{ and } BC = 12\]

(i) \[\sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{12}{13}\]

(ii) \[\cos C = \frac{\text{base}}{\text{hypotenuse}} = \cos C = \frac{BC}{AC} = \frac{12}{13}\]

(iii) \[\tan A = \frac{\text{perpendicular}}{\text{base}} = \frac{BC}{AB} = \frac{12}{5}\]

(iv) \[\cosec C = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{AC}{AB} = \frac{13}{5}\]

(v) Since, \(\sec A = \frac{\text{hypotenuse}}{\text{base}} = \frac{AC}{AB} = \frac{13}{5}\) and \(\tan A = \frac{12}{5}\)

\[\sec^2 A - \tan^2 A = \left(\frac{13}{5}\right)^2 - \left(\frac{12}{5}\right)^2 = \frac{169}{25} - \frac{144}{25} = 1\]

Note: \(\sec^2 A\) means \((\sec A)^2\); \(\tan^2 A\) means \((\tan A)^2\) and so on.

Example 2

In a right-angled triangle, if angle A is acute and \(\cot A = \frac{4}{3}\); find the remaining trigonometrical ratios.

Solution:

Given: \(\cot A = \frac{4}{3}\), i.e., \(\frac{\text{base}}{\text{perpendicular}} = \frac{4}{3}\)

\[\frac{AB}{BC} = \frac{4}{3}\]

If length of AB = 4x unit, length of BC = 3x unit.

Since, \(AC^2 = AB^2 + BC^2\)

\[AC^2 = (4x)^2 + (3x)^2 = 25x^2\]

\[AC = 5x \text{ unit (hyp.)}\]

(i) \[\sin A = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{3x}{5x} = \frac{3}{5}\]

(ii) \[\cos A = \frac{\text{base}}{\text{hypotenuse}} = \frac{4x}{5x} = \frac{4}{5}\]

(iii) \[\tan A = \frac{\text{perpendicular}}{\text{base}} = \frac{3x}{4x} = \frac{3}{4}\]

(iv) \[\sec A = \frac{\text{hypotenuse}}{\text{base}} = \frac{5x}{4x} = \frac{5}{4}\]

(v) \[\cosec A = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{5x}{3x} = \frac{5}{3}\]

Example 3

Given 13 sin A = 12, find:

(i) sec A - tan A (ii) \(\frac{1}{\cos^2 A} - \tan^2 A\)

Solution:

\[13 \sin A = 12 \Rightarrow \sin A = \frac{12}{13}\]

i.e., \[\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{12}{13} \Rightarrow \frac{BC}{AC} = \frac{12}{13}\]

If length of BC = 12x, length of AC = 13x

Since, \(AB^2 + BC^2 = AC^2\)

\[AB^2 + (12x)^2 = (13x)^2\]

\[AB^2 = 169x^2 - 144x^2 = 25x^2\]

\[AB = 5x \text{ (base)}\]

\[\sec A = \frac{\text{hypotenuse}}{\text{base}} = \frac{13x}{5x} = \frac{13}{5}\]

and, \[\tan A = \frac{\text{perpendicular}}{\text{base}} = \frac{12x}{5x} = \frac{12}{5}\]

(i) \[\sec A - \tan A = \frac{13}{5} - \frac{12}{5} = \frac{1}{5}\]

(ii) \[\frac{1}{\cos^2 A} - \tan^2 A = \frac{1}{\left(\frac{5}{13}\right)^2} - \left(\frac{12}{5}\right)^2\]

Note: \(\cos A = \frac{\text{Base}}{\text{Hyp.}} = \frac{5x}{13x}\)

\[= \frac{169}{25} - \frac{144}{25} = \frac{25}{25} = 1\]

Example 4

In the given figure, ABC in a right-angled triangle, right-angled at B. If BC = 5 cm and AC - AB = 1 cm, find the value of cosec A and cos A.

Solution:

BC = 5 cm and AC - AB = 1 cm

\[AC = 1 + AB\]

In right triangle ABC,

\[AC^2 = AB^2 + BC^2\]

\[(1 + AB)^2 = AB^2 + 5^2\]

i.e., \[1 + AB^2 + 2AB = AB^2 + 25\]

\[2AB = 24 \text{ and } AB = 12 \text{ cm}\]

AC - AB = 1

\[AC = 1 + AB = 1 + 12 \text{ cm} = 13 \text{ cm}\]

\[\cosec A = \frac{AC}{BC} = \frac{13}{5} = 2\frac{3}{5} \text{ and } \cos A = \frac{AB}{AC} = \frac{12}{13}\]

Example 5

For the given figure, if \(\cos \alpha = \frac{5}{13}\) and \(\cos \beta = \frac{3}{5}\), find the length of BD.

Solution:

In \(\triangle\) ABC, \[\cos \alpha = \frac{5}{13} \Rightarrow \frac{BC}{AC} = \frac{5}{13}\]

i.e., if BC = 5k, AC = 13k

\[AC^2 = AB^2 + BC^2 \Rightarrow (13k)^2 = (12)^2 + (5k)^2\]

i.e., \[169k^2 = 144 + 25k^2 \Rightarrow 144k^2 = 144\]

i.e., \[k^2 = 1 \text{ and } k = 1\]

\[BC = 5k = 5 \times 1 \text{ m} = 5 \text{ m}\]

In \(\triangle\) CDE, \[\cos \beta = \frac{3}{5} \Rightarrow \frac{CD}{CE} = \frac{3}{5}\]

i.e., if CD = 3p, CE = 5p

\[CE^2 = CD^2 + DE^2 \Rightarrow (5p)^2 = (3p)^2 + (8)^2\]

i.e., \[25p^2 - 9p^2 = 64 \Rightarrow 16p^2 = 64\]

i.e., \[p^2 = 4 \text{ and } p = 2\]

\[CD = 3p = 3 \times 2 \text{ m} = 6 \text{ m}\]

Clearly, \[BD = BC + CD = 5 \text{ m} + 6 \text{ m} = 11 \text{ m}\]

Teacher's Note

Trigonometric ratios help engineers design ramps and staircases by determining the correct angles needed for safe and accessible structures. Understanding these ratios allows professionals to calculate the height and length of stairs required in buildings.

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