Trigonometry, Trigonometric Ratios and Special Angles for SS 1

Class: SS 1
Term: First Term
Week: 5
Age: 15 years
Duration: 60 minutes
Subject: General Mathematics
Curriculum Theme: Number and Numeration
Focal competence: Using trigonometric ratio knowledge to solve angles of elevation and depress 1on
Key competencies/values: Critical Thinking; Collaboration
Skills:

• Using trigonometric ratio knowledge to solve angles of elevation and depression

Previous Lesson: Application of indices, simple indicial equation
Topic: Trigonometry
Subject Matter: Meaning of Trigonometry, Basic trigonometric ratios, Trigonometric ratios of special angles

Specific Objectives

By the end of the lesson, pupils/students should be able to:

Cognitive Domain

• define Trigonometry;
• identify basic trigonometric ratios (sine, cosine, tangent);
• derive trigonometric ratios of special angles 0°, 30°, 45°, 60° and 90°.

Psychomotor Domain

• solve problems involving the use of sine, cosine, and tangent in right-angled triangles;
• apply the use of trigonometric ratios of 30°, 45°, 60° and 90° in solving problems without the use of calculating aids.

Social Domain

• collaborate effectively in group activities to solve problems.

Reference Materials

The following resources were used in planning this lesson:

• 2025 New Revised Senior Secondary Education Curriculum (SSEC)
• Relevant State Unified Scheme of Work
• New General Mathematics for Senior Secondary Schools 1
• The HeadTeacher Scheme of work For The New Revised Senior Secondary Education Curriculum (SSEC)

Instructional Materials

The teacher will teach this lesson with the aid of:

• Chart showing trigonometric ratios of a right-angled triangle
• Pencil and rulers
• Protractor
• Cut out shapes of right-angled triangles showing angles 30°, 45°, 60°
• Chart showing unit circle
• Whiteboard/Blackboard and markers/chalk

Rationale for the Lesson

This lesson is important as trigonometry provides essential tools for solving problems involving angles and distances, which are common in various fields like engineering, physics, and navigation. Understanding basic ratios and special angles forms a foundational knowledge for more advanced mathematical concepts and real-world applications.

Prerequisite/Previous Knowledge

Students should have a basic understanding of angles, types of triangles (especially right-angled triangles), Pythagoras’ theorem, and basic algebraic manipulation.

Lesson Content/Board Summary

Trigonometry

Meaning of Trigonometry

Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. The word “trigonometry” comes from the Greek words “trigonon” (triangle) and “metron” (measure).

It primarily focuses on right-angled triangles and defines six trigonometric ratios (sine, cosine, tangent, cosecant, secant, cotangent), though we will focus on the first three basic ratios.

Basic Trigonometric Ratios

For a right-angled triangle, the basic trigonometric ratios relate the angles to the lengths of its sides. Consider a right-angled triangle ABC, with the right angle at B, and an acute angle (theta) at A.

The sides are named relative to the angle (theta):

1. Opposite (O): The side directly across from angle (theta).
2. Adjacent (A): The side next to angle (theta) that is not the hypotenuse.
3. Hypotenuse (H): The longest side, opposite the right angle.

The three basic trigonometric ratios are:

1. Sine ((sintheta)): Ratio of the length of the opposite side to the length of the hypotenuse.

(sintheta = frac{text{Opposite}}{text{Hypotenuse}} = frac{O}{H})

2. Cosine ((costheta)): Ratio of the length of the adjacent side to the length of the hypotenuse.

(costheta = frac{text{Adjacent}}{text{Hypotenuse}} = frac{A}{H})

3. Tangent ((tantheta)): Ratio of the length of the opposite side to the length of the adjacent side.

(tantheta = frac{text{Opposite}}{text{Adjacent}} = frac{O}{A})

A common mnemonic to remember these ratios is SOH CAH TOA:

• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent

Example 1

Question: In a right-angled triangle PQR, right-angled at Q, if PQ = 3 cm and QR = 4 cm, find (sin P), (cos P), and (tan P).

Solution:

Step 1: Find the length of the hypotenuse PR using Pythagoras’ theorem.

(PR^2 = PQ^2 + QR^2)

(PR^2 = 3^2 + 4^2)

(PR^2 = 9 + 16)

(PR^2 = 25)

(PR = sqrt{25} = 5 text{ cm})

Step 2: Identify the opposite, adjacent, and hypotenuse sides relative to angle P.

Opposite (O) = QR = 4 cm

Adjacent (A) = PQ = 3 cm

Hypotenuse (H) = PR = 5 cm

Step 3: Calculate the trigonometric ratios.

(sin P = frac{O}{H} = frac{4}{5})

(cos P = frac{A}{H} = frac{3}{5})

(tan P = frac{O}{A} = frac{4}{3})

Answer: (sin P = frac{4}{5}), (cos P = frac{3}{5}), (tan P = frac{4}{3})

Trigonometric Ratios of Special Angles (0°, 30°, 45°, 60°, 90°)

These are angles for which the trigonometric ratios can be derived exactly without using a calculator.

Angle 45°

Consider an isosceles right-angled triangle with two equal sides of length 1 unit. The angles are 45°, 45°, and 90°.

Using Pythagoras’ theorem, the hypotenuse (h = sqrt{1^2 + 1^2} = sqrt{2}).

For (theta = 45^circ):

1. (sin 45^circ = frac{text{Opposite}}{text{Hypotenuse}} = frac{1}{sqrt{2}} = frac{sqrt{2}}{2})
2. (cos 45^circ = frac{text{Adjacent}}{text{Hypotenuse}} = frac{1}{sqrt{2}} = frac{sqrt{2}}{2})
3. (tan 45^circ = frac{text{Opposite}}{text{Adjacent}} = frac{1}{1} = 1)

Angles 30° and 60°

Consider an equilateral triangle with side length 2 units. All angles are 60°. If we bisect one angle, we form two right-angled triangles.

Let’s take one of these right-angled triangles. The angles are 30°, 60°, and 90°. The sides are: hypotenuse = 2, adjacent to 60° (opposite to 30°) = 1, and using Pythagoras’ theorem, the other side (opposite to 60°, adjacent to 30°) = (sqrt{2^2 – 1^2} = sqrt{3}).

For (theta = 30^circ):

1. (sin 30^circ = frac{text{Opposite}}{text{Hypotenuse}} = frac{1}{2})
2. (cos 30^circ = frac{text{Adjacent}}{text{Hypotenuse}} = frac{sqrt{3}}{2})
3. (tan 30^circ = frac{text{Opposite}}{text{Adjacent}} = frac{1}{sqrt{3}} = frac{sqrt{3}}{3})

For (theta = 60^circ):

1. (sin 60^circ = frac{text{Opposite}}{text{Hypotenuse}} = frac{sqrt{3}}{2})
2. (cos 60^circ = frac{text{Adjacent}}{text{Hypotenuse}} = frac{1}{2})
3. (tan 60^circ = frac{text{Opposite}}{text{Adjacent}} = frac{sqrt{3}}{1} = sqrt{3})

Angles 0° and 90°

These angles are best understood using the concept of a unit circle or as limiting cases of a right-angled triangle.

Consider a point P(x, y) on a unit circle (radius r=1) with an angle (theta) from the positive x-axis. Then (costheta = x/r = x) and (sintheta = y/r = y). Also, (tantheta = y/x).

For (theta = 0^circ): The point is (1, 0).

1. (sin 0^circ = 0)
2. (cos 0^circ = 1)
3. (tan 0^circ = frac{0}{1} = 0)

For (theta = 90^circ): The point is (0, 1).

1. (sin 90^circ = 1)
2. (cos 90^circ = 0)
3. (tan 90^circ = frac{1}{0}), which is undefined.

Summary Table of Special Angle Ratios

It is useful to memorize these values:

Angle ((theta))	(sintheta)	(costheta)	(tantheta)
0°	0	1	0
30°	(frac{1}{2})	(frac{sqrt{3}}{2})	(frac{1}{sqrt{3}}) or (frac{sqrt{3}}{3})
45°	(frac{1}{sqrt{2}}) or (frac{sqrt{2}}{2})	(frac{1}{sqrt{2}}) or (frac{sqrt{2}}{2})	1
60°	(frac{sqrt{3}}{2})	(frac{1}{2})	(sqrt{3})
90°	1	0	Undefined

Example 2

Question: Evaluate (sin 30^circ + cos 60^circ – tan 45^circ) without using a calculator.

Solution:

Step 1: Recall the values of the trigonometric ratios for the special angles.

(sin 30^circ = frac{1}{2})

(cos 60^circ = frac{1}{2})

(tan 45^circ = 1)

Step 2: Substitute the values into the expression.

(sin 30^circ + cos 60^circ – tan 45^circ = frac{1}{2} + frac{1}{2} – 1)

Step 3: Simplify the expression.

(= 1 – 1 = 0)

Answer: 0

Example 3

Question: If (tan x = sin 60^circ cos 30^circ), find the value of x without using a calculator.

Solution:

Step 1: Recall the values of (sin 60^circ) and (cos 30^circ).

(sin 60^circ = frac{sqrt{3}}{2})

(cos 30^circ = frac{sqrt{3}}{2})

Step 2: Substitute these values into the given equation.

(tan x = left(frac{sqrt{3}}{2}right) times left(frac{sqrt{3}}{2}right))

(tan x = frac{3}{4})

Step 3: This example requires finding an angle whose tangent is 3/4. Since 3/4 is not a standard special angle tangent value, there might be a slight error in the question’s premise for “without using a calculator” if it expects a special angle result. Let’s re-evaluate the question or assume it’s a general application.
If the question intended a special angle, it might have been (tan x = sin 30^circ cos 60^circ) for example, which would give (tan x = frac{1}{2} times frac{1}{2} = frac{1}{4}).
Or, if it was (tan x = sin 60^circ tan 30^circ), then (tan x = frac{sqrt{3}}{2} times frac{1}{sqrt{3}} = frac{1}{2}).
Let’s assume the question meant to lead to a special angle. A common type of problem for special angles is:

Revised Question: If (tan x = frac{sin 60^circ}{cos 60^circ}), find the value of x without using a calculator.

Solution:

Step 1: Recall the values of (sin 60^circ) and (cos 60^circ).

(sin 60^circ = frac{sqrt{3}}{2})

(cos 60^circ = frac{1}{2})

Step 2: Substitute these values into the given equation.

(tan x = frac{frac{sqrt{3}}{2}}{frac{1}{2}})

(tan x = frac{sqrt{3}}{2} times frac{2}{1})

(tan x = sqrt{3})

Step 3: Identify the angle whose tangent is (sqrt{3}) from the special angles table.

From the table, (tan 60^circ = sqrt{3}).

Therefore, (x = 60^circ).

Answer: (x = 60^circ)

Teaching Methods/Instructional Techniques

Discussion, Demonstration, Guided Practice, Question and Answer, Explanation, Group Work, Individual Practice

Instructional Procedures

Step 1: Introduction

Time: 5 minutes

Teaching Skill: Questioning/Recalling

Teacher’s Activity: The teacher greets the students and reviews previous knowledge by asking questions about types of triangles, especially right-angled triangles, and Pythagoras’ theorem. The teacher then introduces the topic of trigonometry as a way to relate angles and sides of triangles.

Pupils’ Activity: Pupils respond to questions about triangles and Pythagoras’ theorem.

Learning Point: Recalling triangle properties

Step 2: Meaning of Trigonometry

Time: 10 minutes

Teaching Skill: Explanation/Definition

Teacher’s Activity: The teacher explains the meaning of trigonometry, its origin, and its focus on relationships between angles and side lengths in triangles, especially right-angled triangles. The teacher uses a chart to illustrate a right-angled triangle and labels its sides relative to an acute angle.

Pupils’ Activity: Pupils listen attentively and ask questions for clarification.

Learning Point: Defining Trigonometry

Step 3: Basic Trigonometric Ratios

Time: 10 minutes

Teaching Skill: Demonstration/Explanation

Teacher’s Activity: The teacher introduces the three basic trigonometric ratios: sine, cosine, and tangent. Using the labeled chart, the teacher defines each ratio in terms of opposite, adjacent, and hypotenuse. The SOH CAH TOA mnemonic is introduced and explained. The teacher works through Example 1 on the board.

Pupils’ Activity: Pupils observe the chart, copy the definitions, and follow the worked example.

Learning Point: Understanding basic ratios

Step 4: Derivation of Special Angles (45°)

Time: 8 minutes

Teaching Skill: Demonstration/Guided Derivation

Teacher’s Activity: The teacher guides students to derive the trigonometric ratios for 45°. This involves drawing an isosceles right-angled triangle (or using a cut-out shape) and applying the definitions of sine, cosine, and tangent. The teacher ensures students understand how the side lengths are determined.

Pupils’ Activity: Pupils participate in the derivation process, drawing the triangle and calculating the ratios.

Learning Point: Deriving 45° ratios

Step 5: Derivation of Special Angles (30°, 60°)

Time: 8 minutes

Teaching Skill: Demonstration/Guided Derivation

Teacher’s Activity: The teacher guides students to derive the trigonometric ratios for 30° and 60°. This involves drawing an equilateral triangle, bisecting it to form a 30-60-90 triangle (or using a cut-out shape), and applying the definitions of sine, cosine, and tangent for both angles.

Pupils’ Activity: Pupils follow the derivation, drawing the triangle and calculating the ratios for 30° and 60°.

Learning Point: Deriving 30° and 60° ratios

Step 6: Special Angles (0°, 90°) and Applications

Time: 9 minutes

Teaching Skill: Explanation/Problem Solving

Teacher’s Activity: The teacher explains the trigonometric ratios for 0° and 90° using the unit circle concept or limiting cases. The teacher then presents the summary table of all special angle ratios and works through Example 2 and the revised Example 3, emphasizing how to use the table to solve problems without calculators. The teacher guides students to work in groups on similar problems, choosing a leader to share findings.

Pupils’ Activity: Pupils listen to the explanation, copy the summary table, and work in groups to solve problems, sharing their findings.

Learning Point: Applying special angle ratios

Step 7: Evaluation/Review

Time: 5 minutes

Teaching Skill: Questioning/Assessment

Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. What is trigonometry?
2. State the basic trigonometric ratios for an acute angle in a right-angled triangle.
3. Derive the value of (sin 45^circ).
4. Evaluate (cos 30^circ + sin 60^circ).

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Assessing ratio understanding

Step 8: Note-Taking

Time: 10 minutes

Teaching Skill: Guided Writing

Teacher’s Activity: The teacher guides pupils/students to copy the essential Board Summary notes, including definitions, basic ratios, derivations, the summary table, and worked examples, into their notebooks.

Pupils’ Activity: Pupils/students copy the notes carefully into their notebooks.

Learning Point: Recording lesson content

Step 9: Conclusion

Time: 5 minutes

Teaching Skill: Summarization

Teacher’s Activity: The teacher briefly summarizes the key concepts of the lesson, reiterating the importance of understanding basic trigonometric ratios and memorizing the special angle values for future mathematical applications. The teacher addresses any final questions.

Pupils’ Activity: Pupils listen and ask any remaining questions.

Learning Point: Consolidating lesson concepts

Continuous Assessment/Further Study

Type: Homework

Instruction: Solve the following problems in your notebook without using a calculator:

1. In a right-angled triangle XYZ, right-angled at Y, if XY = 5 cm and YZ = 12 cm, find (sin X), (cos X), and (tan X).
2. Evaluate (2 sin 30^circ + 3 tan 45^circ).
3. If (tan A = frac{sin 45^circ}{cos 45^circ}), find the value of angle A.
4. A ladder leans against a wall, making an angle of 60° with the ground. If the base of the ladder is 2m from the wall, how long is the ladder? (Hint: Use cosine ratio)

Lesson Keywords

• Trigonometry – The study of relationships between angles and sides of triangles.
• Right-angled triangle – A triangle with one angle measuring 90 degrees.
• Opposite side – The side across from a given angle in a right-angled triangle.
• Adjacent side – The side next to a given angle (not the hypotenuse) in a right-angled triangle.
• Hypotenuse – The longest side of a right-angled triangle, opposite the right angle.
• Sine ((sin)) – Ratio of opposite side to hypotenuse.
• Cosine ((cos)) – Ratio of adjacent side to hypotenuse.
• Tangent ((tan)) – Ratio of opposite side to adjacent side.
• Special angles – Angles (0°, 30°, 45°, 60°, 90°) whose trigonometric ratios can be derived exactly.

Differentiation

For students who grasp the concepts quickly, encourage them to explore the reciprocal trigonometric ratios (cosecant, secant, cotangent) or research simple applications of trigonometry in real-life scenarios. For students needing more support, provide additional practice problems with labeled triangles and guide them through the derivations of special angles step-by-step, focusing on one angle at a time.

Suggested Lesson Videos

Search on YouTube for: “Trigonometric Ratios for SS1” or “Special Angles in Trigonometry Explained”