Lesson Note on Bearing: Cardinal Points and Problem Solving for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: Second Term
Week: 7 – 8
Age: 16 years
Duration: 45 minutes
Subject: General Mathematics
Curriculum Theme: Geometry and Trigonometry
Previous Lesson: .
Topic: BEARING
Subject Matter: Definition and drawing of 4 cardinal 8 cardinal and 16 cardinal points, notation for bearings cardinal notation and three digit notation, sketches involving lengths and angles bearing, problem solving on lengths angles and bearing

Specific Objectives

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

Cognitive Domain:

• Define bearing correctly.
• Identify and draw the 4, 8, and 16 cardinal points.
• Explain and differentiate between cardinal notation and three-digit notation for bearings.
• Sketch diagrams accurately for bearing problems.
• Solve problems involving lengths, angles, and bearings using appropriate mathematical theorems and rules.

Affective Domain:

• Show interest in solving real-world problems involving bearings.
• Appreciate the application of bearing in navigation and surveying.
• Demonstrate carefulness and precision when drawing diagrams and calculating bearings.

Psychomotor Domain:

• Accurately draw and label cardinal points and bearing angles using a protractor and ruler.
• Construct diagrams from given bearing problems.
• Apply Pythagoras theorem, trigonometric ratios, sine rule, and cosine rule effectively to calculate unknown distances and angles.

Social Domain:

• Collaborate with peers to discuss and solve bearing-related problems.
• Communicate their problem-solving steps clearly to others.

Reference Materials

The following resources were used in planning this lesson:

• 9 Years Basic Education Curriculum (Mathematics for Senior Secondary Schools)
• State Unified Scheme of Work for Senior Secondary Schools
• New General Mathematics for Senior Secondary Schools 2

Instructional Materials

The teacher will teach this lesson with the aid of:

• Charts illustrating cardinal points
• Ruler
• Pencil
• Protractor
• Computer assisted resources (if available)

Rationale for the Lesson

Understanding bearing helps pupils to determine direction and position, which is important in navigation, surveying, and geography. This knowledge enables pupils to solve practical problems related to movement and location in their daily lives and future careers.

Prerequisite/Previous Knowledge

Pupils have prior knowledge of angles, basic compass directions, properties of parallel lines, Pythagoras theorem, and basic trigonometric ratios from Junior Secondary School.

Lesson Content/Board Summary

Bearing

Definition of Bearing

Bearing is the direction of one point from another, usually measured as an angle from the North line. It helps in locating positions and charting courses.

Cardinal Points

Cardinal points are the principal directions of a compass. They serve as fundamental reference points for determining bearings.

The following are the cardinal points:

• 4 Cardinal Points: North (N), East (E), South (S), West (W).
• 8 Cardinal Points: North (N), North-East (NE), East (E), South-East (SE), South (S), South-West (SW), West (W), North-West (NW). These are obtained by bisecting the angles between the 4 cardinal points.
• 16 Cardinal Points: These are obtained by bisecting the angles between the 8 cardinal points. Examples include North-North-East (NNE), East-North-East (ENE), East-South-East (ESE), South-South-East (SSE), South-South-West (SSW), West-South-West (WSW), West-North-West (WNW), North-North-West (NNW).

Notation for Bearings

There are two main ways to express bearings:

1. Cardinal Notation (Compass Bearing)

This notation uses the cardinal points (N, S, E, W) to describe a direction. The angle is always acute (less than 90°) and is measured from either the North or South line towards East or West.

Examples:

• N30°E: 30 degrees East of North.
• S60°W: 60 degrees West of South.
• N45°W: 45 degrees West of North (same as NW).

2. Three-Digit Notation (True Bearing)

This notation measures the angle clockwise from the North line. The angle is always written using three digits, even if the angle is less than 100 degrees (e.g., 045°). It ranges from 000° to 360°.

Examples:

• 045°: This is equivalent to N45°E or NE.
• 120°: This is measured 120° clockwise from North.
• 270°: This is measured 270° clockwise from North (due West).
• 090°: This is measured 90° clockwise from North (due East).

Note: When converting from cardinal to three-digit notation, always remember to add to 90°, 180°, or 270° depending on the quadrant.

Sketching Bearing Problems

Accurate diagrams are essential for solving bearing problems. The following steps guide sketching:

• Draw a North line at the starting point of the bearing.
• Measure the angle clockwise from the North line for three-digit bearings, or from North/South towards East/West for cardinal bearings.
• Draw a line representing the path or direction of travel.
• Mark the destination point and draw another North line at this point if a new bearing is given from there.
• Label all known lengths and angles clearly.

Problem Solving in Bearing

Solving bearing problems often involves forming triangles and applying geometric principles along with trigonometric rules.

1. Using Pythagoras Theorem

Pythagoras theorem is used for right-angled triangles to find unknown side lengths. The formula is a² + b² = c², where ‘c’ is the hypotenuse.

Example 1: A ship sails 8 km due East from point A to point B, then 6 km due North from point B to point C. Calculate the distance from A to C.

Solution:

1. Step 1: Sketch the diagram. Draw point A, then a line 8 km East to B. From B, draw a line 6 km North to C. This forms a right-angled triangle ABC, with the right angle at B.
2. Step 2: Identify the sides. AB = 8 km, BC = 6 km. AC is the hypotenuse.
3. Step 3: Apply Pythagoras theorem. AC² = AB² + BC²
4. Step 4: Substitute values and solve. AC² = 8² + 6² = 64 + 36 = 100
5. Step 5: Find the square root. AC = √100 = 10 km.

The distance from A to C is 10 km.

Example 2: A plane flies 120 km North from P to Q. It then flies West to R. If the distance from P to R is 150 km, how far did the plane fly from Q to R?

Solution:

1. Step 1: Sketch the diagram. Draw P, then 120 km North to Q. From Q, draw a line West to R. Connect P to R (150 km). This forms a right-angled triangle PQR, with the right angle at Q.
2. Step 2: Identify the sides. PQ = 120 km, PR = 150 km (hypotenuse). QR is the unknown side.
3. Step 3: Apply Pythagoras theorem. PR² = PQ² + QR²
4. Step 4: Substitute values. 150² = 120² + QR²
5. Step 5: Isolate QR². 22500 = 14400 + QR² => QR² = 22500 – 14400 = 8100
6. Step 6: Find the square root. QR = √8100 = 90 km.

The plane flew 90 km from Q to R.

2. Using Trigonometric Ratios (SOH CAH TOA)

These ratios (sine, cosine, tangent) are used for right-angled triangles to find unknown angles or sides when one angle and one side, or two sides, are known.

• Sine (sin θ) = Opposite / Hypotenuse
• Cosine (cos θ) = Adjacent / Hypotenuse
• Tangent (tan θ) = Opposite / Adjacent

Example 1: A ladder 10 m long leans against a vertical wall. If the foot of the ladder is 6 m from the base of the wall, calculate the angle the ladder makes with the ground.

Solution:

1. Step 1: Sketch the diagram. Draw a right-angled triangle where the ladder is the hypotenuse (10 m), the distance from the wall is the adjacent side (6 m), and the angle with the ground is θ.
2. Step 2: Identify knowns and unknowns. Hypotenuse = 10 m, Adjacent = 6 m. We need to find θ.
3. Step 3: Choose the appropriate ratio. Since we have Adjacent and Hypotenuse, use Cosine. cos θ = Adjacent / Hypotenuse.
4. Step 4: Substitute values. cos θ = 6 / 10 = 0.6
5. Step 5: Find the angle. θ = cos⁻¹(0.6) ≈ 53.13° (to 2 decimal places).

The ladder makes an angle of approximately 53.13° with the ground.

Example 2: From a point P, a tower is observed at an angle of elevation of 35°. If the point P is 50 m from the base of the tower, calculate the height of the tower.

Solution:

1. Step 1: Sketch the diagram. Draw a right-angled triangle. The height of the tower is the Opposite side, the distance from the base is the Adjacent side (50 m), and the angle of elevation is 35°.
2. Step 2: Identify knowns and unknowns. Angle θ = 35°, Adjacent = 50 m. We need to find Opposite (height, h).
3. Step 3: Choose the appropriate ratio. Since we have Opposite and Adjacent, use Tangent. tan θ = Opposite / Adjacent.
4. Step 4: Substitute values. tan 35° = h / 50
5. Step 5: Solve for h. h = 50 × tan 35°
6. Step 6: Calculate the value. h ≈ 50 × 0.7002 ≈ 35.01 m (to 2 decimal places).

The height of the tower is approximately 35.01 m.

3. Using Sine Rule

The Sine Rule is used for any triangle (not just right-angled) when you know:

• Two angles and one side (AAS or ASA).
• Two sides and an angle opposite one of them (SSA – ambiguous case).

The formula is: a/sin A = b/sin B = c/sin C

Example 1: In triangle ABC, angle A = 40°, angle B = 60°, and side a = 8 cm. Find side b.

Solution:

1. Step 1: Identify knowns. A = 40°, B = 60°, a = 8 cm. We need to find b.
2. Step 2: Apply the Sine Rule. a/sin A = b/sin B
3. Step 3: Substitute values. 8/sin 40° = b/sin 60°
4. Step 4: Solve for b. b = (8 × sin 60°) / sin 40°
5. Step 5: Calculate the value. b ≈ (8 × 0.8660) / 0.6428 ≈ 6.928 / 0.6428 ≈ 10.78 cm (to 2 decimal places).

Side b is approximately 10.78 cm.

Example 2: A hiker walks 5 km from A to B on a bearing of 060°. From B, he walks to C. The bearing of C from A is 090°, and the bearing of C from B is 150°. Find the distance BC.

Solution:

1. Step 1: Sketch the diagram and find angles.
   • Draw North lines at A and B.
   • Bearing of B from A is 060°, so angle NAB = 60°.
   • Bearing of C from A is 090°, so angle NAC = 90°. Angle BAC = 90° – 60° = 30°.
   • Bearing of C from B is 150°. The interior angle ABC can be found. Draw a North line at B, the angle from North to BC is 150°. The line AB makes an angle of 60° with the North line at A. Due to parallel North lines, the angle NBA (alternate interior angle) is 60°. Angle N’BC (where N’ is North at B) is 150°. Angle ABC = (180° – 60°) + (180° – 150°) = 120° + 30° = 150°. More simply, angle N’BA = 60° (alternate to NAB). Angle ABC = 180° – (180° – 150°) – 60° = 180° – 30° – 60° = 90°. (Angle N’BC = 150°. Angle between N’B and BA is 60°. So angle ABC = 150 – 60 = 90. This is incorrect. Angle NBA = 60. Angle N’BC = 150. Angle ABC = 180 – (angle N’BC – angle N’BA) = 180 – (150-60) = 180 – 90 = 90. This is also incorrect.
     Let’s re-evaluate angle ABC carefully:
     The North line at A and the North line at B are parallel.
     Angle NAB = 60°. Therefore, the interior angle at B, formed by line AB and the North line *extending South from B*, is also 60° (alternate angles).
     The bearing of C from B is 150°. This means the angle measured clockwise from the North line at B to the line BC is 150°.
     The angle between the North line at B and the line BA (measured clockwise from North) is 180° + 60° = 240° (back bearing of A from B is 240°).
     Angle ABC = Bearing of C from B – (Back bearing of A from B – 180). This is getting complicated.
     Let’s use angles in a triangle:
     Angle A = 90 – 60 = 30°.
     Angle at B: Angle N’BA = 60° (alternate to NAB). The bearing of C from B is 150°. So, the angle between the North line N’B and BC is 150°.
     Angle ABC = 360° – (Bearing of C from B) – (Angle between BA and North line at B).
     Angle N’BA = 60°.
     Angle ABC = 180° – (150° – 60°) = 180° – 90° = 90°. This is correct.
     So, Angle ABC = 90°.
     Angle C = 180° – 30° – 90° = 60°.
2. Step 2: Apply Sine Rule. We have side a (BC), angle A (30°), side c (AB=5km), angle C (60°).
   We want to find BC (side a). So, a/sin A = c/sin C.
3. Step 3: Substitute values. BC/sin 30° = 5/sin 60°
4. Step 4: Solve for BC. BC = (5 × sin 30°) / sin 60°
5. Step 5: Calculate the value. BC = (5 × 0.5) / 0.8660 ≈ 2.5 / 0.8660 ≈ 2.89 km (to 2 decimal places).

The distance BC is approximately 2.89 km.

4. Using Cosine Rule

The Cosine Rule is used for any triangle when you know:

• Two sides and the included angle (SAS).
• All three sides (SSS) to find an angle.

The formulas are:

• a² = b² + c² – 2bc cos A
• b² = a² + c² – 2ac cos B
• c² = a² + b² – 2ab cos C

Example 1: Two ships leave a port P. Ship A sails 60 km on a bearing of 040°. Ship B sails 80 km on a bearing of 120°. How far apart are the two ships?

Solution:

1. Step 1: Sketch the diagram and find the included angle.
   • Draw port P with a North line.
   • Draw PA (60 km) at 040° from North.
   • Draw PB (80 km) at 120° from North.
   • The angle APB (included angle) = Bearing of B – Bearing of A = 120° – 40° = 80°.
   • We have two sides (PA=60, PB=80) and the included angle (APB=80°). This is SAS.
2. Step 2: Apply the Cosine Rule. Let the distance between ships be AB (side p).
   p² = a² + b² – 2ab cos P (using P for angle APB, a for PB, b for PA)
   AB² = PA² + PB² – 2(PA)(PB) cos(APB)
3. Step 3: Substitute values. AB² = 60² + 80² – 2(60)(80) cos 80°
4. Step 4: Calculate. AB² = 3600 + 6400 – 9600 × 0.1736
5. Step 5: Simplify. AB² = 10000 – 1666.56 = 8333.44
6. Step 6: Find AB. AB = √8333.44 ≈ 91.29 km (to 2 decimal places).

The two ships are approximately 91.29 km apart.

Example 2: In triangle ABC, a = 7 cm, b = 9 cm, c = 12 cm. Find angle A.

Solution:

1. Step 1: Identify knowns. a = 7, b = 9, c = 12. We need to find angle A.
2. Step 2: Apply the Cosine Rule for angle A. a² = b² + c² – 2bc cos A.
   Rearrange to find cos A: cos A = (b² + c² – a²) / (2bc)
3. Step 3: Substitute values. cos A = (9² + 12² – 7²) / (2 × 9 × 12)
4. Step 4: Calculate. cos A = (81 + 144 – 49) / 216 = (225 – 49) / 216 = 176 / 216
5. Step 5: Find the angle. cos A ≈ 0.8148. A = cos⁻¹(0.8148) ≈ 35.43° (to 2 decimal places).

Angle A is approximately 35.43°.

Finding Back Bearings:

The bearing of point A from point B is called the back bearing of B from A. If the bearing of B from A is θ:

• If θ < 180°, the back bearing is θ + 180°.
• If θ > 180°, the back bearing is θ – 180°.

Example: The bearing of a town B from town A is 070°. What is the bearing of town A from town B?

Solution: Since 070° < 180°, the back bearing = 070° + 180° = 250°.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes
Teaching Skill: Set Induction and Questioning

Teacher’s Activities:
Revises angles and compass directions by asking brief questions. Introduces the topic Bearing and states lesson objectives.

Pupils’ Activities:
Respond to questions and mention examples where direction is used.

Learning Point:
Direction can be described using bearings.

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Step 2: Cardinal Points and Meaning of Bearing

Time: 8 minutes
Teaching Skill: Explanation and Demonstration

Teacher’s Activities:
Explains bearing and demonstrates the 4, 8, and 16 cardinal points using charts and board sketches.

Pupils’ Activities:
Observe, copy diagrams, and label cardinal points.

Learning Point:
Bearings are measured clockwise from North using cardinal points.

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Step 3: Bearing Notations

Time: 7 minutes
Teaching Skill: Guided Instruction

Teacher’s Activities:
Explains cardinal notation and three-digit notation with examples. Demonstrates conversions.

Pupils’ Activities:
Follow examples and practice simple conversions.

Learning Point:
Bearings can be written in compass form or three-digit form.

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Step 4: Sketching Bearings

Time: 7 minutes
Teaching Skill: Demonstration

Teacher’s Activities:
Demonstrates how to sketch bearing diagrams using North lines, angles, and labeled points.

Pupils’ Activities:
Construct sample diagrams using rulers and protractors.

Learning Point:
Accurate sketches are necessary for solving bearing problems.

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Step 5: Solving Bearing Problems

Time: 10 minutes
Teaching Skill: Guided Practice

Teacher’s Activities:
Works through examples involving distances and angles using Pythagoras theorem, trigonometric ratios, sine rule, and cosine rule.

Pupils’ Activities:
Solve similar problems individually and in pairs.

Learning Point:
Bearings form triangles that can be solved using appropriate mathematical rules.

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Step 6: Evaluation

Time: 5 minutes
Teaching Skill: Questioning

Teacher’s Activities:
Asks pupils to:

1. Define bearing.

2. State cardinal points.

3. Convert a bearing.

4. Solve a short problem.

Pupils’ Activities:
Respond orally and in writing.

Learning Point:
Pupils demonstrate understanding of bearings.

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Step 7: Conclusion

Time: 3 minutes

Teacher’s Activities:
Summarizes key points and gives classwork/homework.

Pupils’ Activities:
Listen and copy assignment.

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Lesson Keywords

• Bearing – The direction of one point from another measured clockwise from North.
• Cardinal Points – Main compass directions used for navigation (N, E, S, W and their subdivisions).
• True Bearing – A three-digit angle measured clockwise from North.
• Back Bearing – The bearing of the starting point from the destination.
• Protractor – An instrument used to measure angles.
• Sketch – A simple diagram showing directions, distances, and angles.

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Differentiation

Fast learners solve extended bearing problems involving sine and cosine rules.
Average learners practice standard problems with teacher guidance.
Slow learners receive additional support with diagram construction and basic bearing conversions.

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Note to Teachers

Emphasize accurate diagram drawing and correct use of protractors.
Encourage pupils to always draw North lines before attempting bearing questions.
Relate examples to real-life situations such as navigation and map reading to improve understanding.
Monitor pupils closely during calculations to reduce errors in angle measurement and trigonometric application.