# Class-9 Trigonometry

Trigonometrical ratios of angles of 30^{o} and 60^{o}

Trigonometrical ratios of angles of 45^{o}

## Introduction to Trigonometry

Measurement of triangles is known as Trigonometry. Here we shall be dealing with the relation of sides and angles of a right-angled triangle.

In the above diagram Base = AB, Perpendicular = BC, Hypotenuse = AC and angle of reference = ∠A

## Trigonometrical Ratios

The ratio between the lengths of two sides of a right-angled triangle is known as trigonometrical ratios. There are six trigonometrical ratios present for a right-angled triangle, they are given below.

Sin A = ^{Perpendicular}⁄_{Hypotenuse} = ^{BC}⁄_{AC}

Cos A = ^{Base}⁄_{Hypotenuse} = ^{AB}⁄_{AC}

Tan A = ^{Perpendicular}⁄_{Base} = ^{BC}⁄_{AB}

Cot A = ^{Base}⁄_{Perpendicular} = ^{AB}⁄_{BC}

Sec A = ^{Hypotenuse}⁄_{Base} = ^{AC}⁄_{AB}

Cosec A = ^{Hypotenuse}⁄_{Perpendicular} = ^{AC}⁄_{BC}

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

**Example 1.** From the below given figure find the value of Sin A, Cos A and Tan A.

**Solution.** As we can see, above figure is a wright angled triangle.

=> AC^{2} = AB^{2} + BC^{2}

=> 13^{2} = 52 + BC^{2}

=> BC^{2} = 169 − 25

=> BC^{2} = 144

=> BC = √ 144

=> BC = 12

Sin A = ^{BC}⁄_{AC} = ^{12}⁄_{13}

Cos A = ^{AB}⁄_{AC} = ^{5}⁄_{13}

Tan A = ^{BC}⁄_{AB} = ^{12}⁄_{5}

**Example 2.** In a right-angled triangle, if Tan A = 4/3 then find the value of Sin a, Cos A, Cot A, Sec A and Cosec A.

**Solution.**

Tan A = ^{4}⁄_{3}

=> Perpendicular/Base = ^{4}⁄_{3}

=> ^{BC}⁄_{AB} = ^{4}⁄_{3}

Let us assume AB = 3x and BC = 4x

AC^{2} = AB^{2} + BC^{2}

=> AC^{2} = (3x)^{2} + (4x)^{2}

=> AC^{2} = 9x^{2} + 16x^{2}

=> AC = √ 25x^{2}

=> AC = 5x

Sin A = ^{4x}⁄_{5x} = ^{4}⁄_{5}

Cos A = ^{3x}⁄_{5x} = ^{3}⁄_{5}

Cot A = ^{3x}⁄_{4x} = ^{3}⁄_{4}

Cosec A = ^{5x}⁄_{4x} = ^{5}⁄_{4}

Sec A = ^{5x}⁄_{3x} = ^{5}⁄_{3}

**Example 3.** If Cos A = ^{4x}⁄_{5x}, find the value of Tan A − Sec A.

**Solution.**

Let's assume AB = 5x and AC = 13x.

AB^{2} + BC^{2} = AC^{2}

=> BC^{2} = AC^{2} &miuns; AB^{2}

=> BC^{2} = (13x)^{2} − (5x)^{2}

=> BC^{2} = 169x^{2} − 25x^{2}

=> BC = √ 144x^{2}

=> BC = 12x

Tan A − Sec A

=> ^{BC}⁄_{AB} − ^{AC}⁄_{AB}

=> ^{12}⁄_{5} − ^{13}⁄_{5}

=> ^{(12−13)}⁄_{5}

=> ^{-1}⁄_{5}

**Example 4.** In the below given figure, ABC is a right-angled triangle. If BC = 5 cm and AC − BC = 1 cm, then find the value of Sin A and Sec A.

**Solution.** Here AB = 5 cm and AC − BC = 1 cm.

AC = 1 + BC

As we know, in right angle triangle,

AC^{2} = AB^{2} + BC^{2}

=> (1 + BC)^{2} = 25 + BC^{2}

=> 1 + BC^{2} + 2BC = 25 + BC^{2}

=> 2BC = 24

=> BC = 12 cm

AC − BC = 1

=> AC = 13 cm

Sin A = ^{BC}⁄_{AC} = ^{12}⁄_{13}

Sec A = ^{AC}⁄_{AB} = ^{13}⁄_{5}

**Example 5.** In the below given triangle ABC, AD is perpendicular to BC. If BC = 112 cm, Tan B = 3/4 and Tan C = 5/12 . Find the length of BP and PC.

**Solution.** As given Tan B = ^{3}⁄_{4}

=> ^{AP}⁄_{BP} = ^{3}⁄_{4}

We can assume AP = 3x and BP = 4x

Similarly, Tan C = ^{5}⁄_{12}

=> ^{AP}⁄_{PC} = ^{5}⁄_{12}

Here, we can assume AP = 5x and PC = 12x

BP + PC = 112

=> 4x + 12x = 112

=> 16x = 112

=> x = ^{112}⁄_{16}

=> x = 7

BP = 4x = 4 × 7 = 28 cm

PC = 12x = 12 × 7 = 84 cm

## Reciprocal Relations

Sin A = ^{Perpendicular}⁄_{Hypotenuse} and Cosec A = ^{Hypotenuse}⁄_{Perpendicular}

As we can see Sin A and Cosec A are reciprocal to each other.

Hence, Sin A = ^{1}⁄_{Cosec A} and Cosec A = ^{1}⁄_{Sin A}

Cos A = ^{Base}⁄_{Hypotenuse} and Sec A = ^{Hypotenuse}⁄_{Base}

Hence, Cos A = ^{1}⁄_{sec A} and Sec A = ^{1}⁄_{Cos A}

Tan A = ^{Perpendicular}⁄_{Base} and Cot A = ^{Base}⁄_{Perpendicular}

Here, we can see Tan A and Cot A are reciprocal to each other.

Hence Tan A = ^{1}⁄_{Cot A} and Cot A = ^{1}⁄_{Tan A}

## Trigonometrical ratios of angles of 30^{o} and 60^{o}

In the below given diagram, lets assume ABC is an equilateral triangle having each side as 2a and AP is perpendicular to BC.

Here BP = PC = a
In right angle triangle AB^{2} = BP^{2} + AP^{2}

=> AP^{2} = AB^{2} − BP^{2}

=> AP^{2} = (2a)^{2} − a^{2}

=> AP^{2} = 4a^{2} − a^{2}

=> AP^{2} = 3a^{2}

=> AP = √3a

Sin 60^{o} = ^{√3a}⁄_{2a} = ^{√3}⁄_{2}

Sin 30^{o} = ^{a}⁄_{2a} = ^{1}⁄_{2}

Cos 60^{o} = ^{a}⁄_{2a} = ^{1}⁄_{2}

Cos 30^{o} = ^{√3a}⁄_{2a} = ^{√3}⁄_{2}

Tan 60^{o} = ^{√3a}⁄_{a} = √ 3

Tan 30^{o} = ^{a}⁄_{√3a} = ^{1}⁄_{√3}

## Trigonometrical ratios of angles of 45^{o}

Below given diagram is a right-angled isosceles triangle in which ∠B = 90^{o} and AB = BC = a

Here ∠A = 45^{o}

Sin 45^{o} = ^{a}⁄_{√2a} = ^{1}⁄_{√2}

Cos 45^{o} = ^{a}⁄_{√2a} = ^{1}⁄_{√2}

Tan 45^{o} = ^{a}⁄_{a} = 1

Angle | 0^{o} |
30^{o} |
45^{o} |
60^{o} |
90^{o} |
---|---|---|---|---|---|

Sin | 0 | ^{1}⁄_{2} |
^{1}⁄_{√2} |
^{√3}⁄_{2} |
1 |

Cos | 1 | ^{√3}⁄_{2} |
^{1}⁄_{√2} |
^{1}⁄_{2} |
0 |

Tan | 0 | ^{1}⁄_{√3} |
1 | √ 3 | Not Defined |

Cot | Not Defined | √ 3 | 1 | ^{1}⁄_{√3} |
0 |

Sec | 1 | ^{2}⁄_{√3} |
√ 2 | 2 | Not Defined |

Cosec | Not Defined | 2 | √ 2 | ^{2}⁄_{√3} |
1 |

From the above table, we can observe that as the angle increases from 0^{o} to 90^{o}

- Value of Sin increases from 0 to 1
- Value of Cos decreases from 1 to 0
- Value of Tan increases from 0 to infinite

Few formulas to remember for future problem solving.

- Sin
^{2}A + Cos^{2}A = 1 - Sec
^{2}A − Tan^{2}A = 1 - Cosec
^{2}A − Cot^{2}A = 1

**Example 1.** Find the value of Sin^{2}30^{o} + 4 Cos^{2}90^{o} + 5 Tan^{3}45^{o}.

**Solution.** Sin^{2}30^{o} + 4 Cos^{2}90^{o} + 5 Tan^{3}45^{o}

= (^{1}⁄_{2})^{2} + 4 × (0)^{2} + 5 × (1)^{3}

= ^{1}⁄_{4} + 0 + 5

= ^{21}⁄_{4}

**Example 2.** If A = 45^{o}, then verify Sec^{2}A − Tan^{2}A = 1.

**Solution.** Sec^{2}45^{o} − Tan^{2}45^{o}

= (√ 2 )^{2} − (1)^{2}

= 2 − 1

= 1

**Example 3.** If a = 15^{o}, then find the value of 4 Sin 2a × Cos 3a × Cot 6a.

**Solution.** 4 Sin 2a × Cos 3a × Cot 6a

= 4 Sin 30^{o} × Cos 45^{o} × Cot 90^{o}

= 4 × ^{1}⁄_{2} × ^{1}⁄_{√2} × 0

= 0

**Example 4.** Prove that Sin 60^{o} = 2 ^{Tan 30o}⁄_{(1+Tan230o)}.

**Solution.** First consider LHS (Left Hand Side)

Sin 60^{o} = ^{√3}⁄_{2}

Now, lets consider RHS (Right Hand Side)

2 ^{Tan 30o}⁄_{(1+Tan230o)}

= 2 (^{1⁄√3}⁄_{1+1⁄3})

= ^{2⁄√3}⁄_{4⁄3}

= ^{2}⁄_{√3} × ^{3}⁄_{4}

= ^{√ 3 }⁄_{2}

Hence LHS = RHS.

## Class-9 Trigonometry MCQ

## Class-9 Trigonometry Worksheets

## Answer Sheet

**Trigonometry-Answer**Download the pdf

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