Class-9 Trigonometry

Introduction to Trigonometry

Trigonometrical Ratios

Reciprocal Relations

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

Trigonometrical ratios of angles of 45^o

Trigonometrical MCQ

Trigonometry Worksheets

Answer Sheet

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.

Class-9-Trigonometry
Solution. As we can see, above figure is a wright angled triangle.

        => AC² = AB² + BC²

        => 13² = 52 + BC²

        => BC² = 169 − 25

        => BC² = 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.
Class-9-Trigonometry
Tan A = ⁴⁄₃

=> Perpendicular/Base = ⁴⁄₃

=> ^BC⁄_AB = ⁴⁄₃

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

AC² = AB² + BC²

=> AC² = (3x)² + (4x)²

=> AC² = 9x² + 16x²

=> AC = √ 25x²

=> 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.
Class-9-Trigonometry
Let's assume AB = 5x and AC = 13x.

AB² + BC² = AC²

=> BC² = AC² &miuns; AB²

=> BC² = (13x)² − (5x)²

=> BC² = 169x² − 25x²

=> BC = √ 144x²

=> 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.

Class-9-Trigonometry
Solution. Here AB = 5 cm and AC − BC = 1 cm.

AC = 1 + BC

As we know, in right angle triangle,

AC² = AB² + BC²

=> (1 + BC)² = 25 + BC²

=> 1 + BC² + 2BC = 25 + BC²

=> 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.

Class-9-Trigonometry
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.
Class-9-Trigonometry
Here BP = PC = a In right angle triangle AB² = BP² + AP²

=> AP² = AB² − BP²

=> AP² = (2a)² − a²

=> AP² = 4a² − a²

=> AP² = 3a²

=> 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
Class-9-Trigonometry
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.

Sin²A + Cos²A = 1
Sec²A − Tan²A = 1
Cosec²A − Cot²A = 1

Example 1. Find the value of Sin²30^o + 4 Cos²90^o + 5 Tan³45^o.

Solution. Sin²30^o + 4 Cos²90^o + 5 Tan³45^o

= (¹⁄₂)² + 4 × (0)² + 5 × (1)³

= ¹⁄₄ + 0 + 5

= ²¹⁄₄

Example 2. If A = 45^o, then verify Sec²A − Tan²A = 1.

Solution. Sec²45^o − Tan²45^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 30^o}⁄_{(1+Tan²30^o)}.

Solution. First consider LHS (Left Hand Side)

Sin 60^o = ^√3⁄₂

Now, lets consider RHS (Right Hand Side)

2 ^{Tan 30^o}⁄_{(1+Tan²30^o)}

= 2 (^¹⁄_√3⁄_1+¹⁄₃)

= ^²⁄_√3⁄_⁴⁄₃

= ²⁄_√3 × ³⁄₄

= ^√ 3⁄₂

Hence LHS = RHS.