IGCSE Mathematics: Geometry and Trigonometry — Bearings, Pythagoras, Sine and Cosine Rules

TL;DR

This topic combines essential geometry and trigonometry skills to solve problems involving distances and directions. You'll learn to use bearings for direction, Pythagoras' theorem for right-angled triangles, and the Sine/Cosine Rules for non-right-angled triangles. Mastering these tools lets you find unknown sides or angles in various scenarios.

1. The Mental Model

Think of this topic as having a toolbox for measuring and locating things on a map. Bearings tell you which way to look, Pythagoras helps when you have a perfect corner, and the Sine/Cosine Rules are your go-to for all other triangle shapes.

2. The Core Material

Bearings

Bearings are used to describe a direction from one point to another. They have three key rules:
1. Measured from North: Always start measuring from the North line, which points straight up.
2. Clockwise: Always measure the angle in a clockwise direction.
3. Three Figures: Always write the bearing using three figures (e.g., 045°, 120°, 270°).

Remember that North lines are parallel. This is crucial for using properties of parallel lines (like alternate angles) when solving problems.

Pythagoras' Theorem

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Pythagoras' Theorem applies only to right-angled triangles. It states that the square of the hypotenuse (the longest side, opposite the right angle) is equal to the sum of the squares of the other two sides.
$a^2 + b^2 = c^2$
Here, 'c' is always the hypotenuse. You'll use this to find an unknown side when you know the other two sides.

Sine and Cosine Rules

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These rules are for non-right-angled triangles, though they work for right-angled ones too.

Sine Rule

You use the Sine Rule when you have:
* Two angles and one side (AAS or ASA)
* Two sides and a non-included angle (SSA – be careful, this can sometimes give two possible triangles)

The rule is:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Where 'a' is the side opposite angle 'A', 'b' opposite 'B', and 'c' opposite 'C'. You pick the pair you know and the pair you want to find.

Cosine Rule

You use the Cosine Rule when you have:
* Two sides and the included angle (SAS) – to find the third side.
* All three sides (SSS) – to find any angle.

To find a side:
$a^2 = b^2 + c^2 - 2bc \cos A$

To find an angle (rearranged):
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

When to use which rule?

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Here's a simple flowchart to help you decide:

graph TD
    A["Start"] --> B{Is it a right-angled triangle?};
    B -- Yes --> C{Do you know two sides and want the third?};
    C -- Yes --> D["Use Pythagoras' Theorem: a² + b² = c²"];
    C -- No --> E{Do you know an angle and a side, or want to find one?};
    E -- Yes --> F["Use SOH CAH TOA (basic trigonometry)"];
    B -- No --> G{Do you have two sides and the angle between them (SAS) or all three sides (SSS)?};
    G -- Yes --> H["Use the Cosine Rule"];
    G -- No --> I{Do you have a side and its opposite angle, plus one other side or angle?};
    I -- Yes --> J["Use the Sine Rule"];
    D --> K["End"];
    F --> K;
    H --> K;
    J --> K;

3. Worked Example

A ship sails 8 km from Port A to Port B on a bearing of 060°. It then sails 6 km from Port B to Port C on a bearing of 150°. Calculate the distance AC and the bearing of C from A.

1. Sketch the situation: Draw North lines at A and B. Mark the bearings and distances.

2. Find angle ABC:

   • Bearing of B from A is 060°.
   • The North line at A and the North line at B are parallel. The line AB is a transversal. So, the angle inside the triangle at B, from the North line pointing south (opposite the 060°), is also 060° (alternate interior angles).
   • The bearing of C from B is 150°.
   • Angle ABC = (360° - 150°) + 60° = 210° (reflex angle outside triangle)
   • No, that's not right. Let's look at it differently: Angle North-B-A is 60°. Angle North-B-C is 150°. The angle between the path from A and the path to C (angle ABC) is the angle from the line BA to the line BC. The angle from the line AB to the North line at B (the back bearing from A) is 060° + 180° = 240°. No, that's not helpful either.

   Let's use the internal angles with the parallel North lines.
   * Draw the North line at B. The angle from the North line (pointing down from B, parallel to AB) to the line BA is 60° (alternate angles with the 060° bearing from A).
   * The bearing of BC is 150°. This means the angle from the North line up from B, clockwise to BC, is 150°.
   * The angle between the North line pointing down from B and the North line pointing up from B is 180°.
   * So, the angle from the line BA to the North line pointing up from B is 180° - 60° = 120°.
   * Therefore, angle ABC = 120° (from BA to North up) + (Angle from North up to BC which is 150°) = No.

   Let's try again with the common technique.
   * Draw North line at B. The angle from North at B clockwise to BA (the back bearing to A) is 060° + 180° = 240°.
   * The angle North-B-C is 150°.
   * Angle ABC = 240° - 150° = 90°. Wait, that's assuming the 240 is measured from North to BA.

   Let's redraw and use simpler angles:
   * From A, bearing to B is 060°.
   * At B, draw a North line. The line BA makes an angle of 60° with the South line from B (alternate interior angles with the 60° from A).
   * The bearing to C from B is 150°. This means the angle from North at B clockwise to BC is 150°.
   * The angle from the South line at B clockwise to BC would be 150° - 180° = -30° (or 360-150-180). This isn't right.
   * The angle from the North line (pointing up from B) counter-clockwise to BA is 60°.
   * The angle from the North line (pointing up from B) clockwise to BC is 150°.
   * The angle inside the triangle at B (angle ABC) is the angle between BA and BC.
   * Angle North-B-A is 180° - 60° = 120° (interior angles between parallel lines).
   * Angle ABC = Angle North-B-C - Angle North-B-A = 150° - (180°-60°) = 150° - 120° = 30°. This is correct! (The angle between the line BA and the North line at B is 180-60=120. The angle between the North line and BC is 150. So 150-120 = 30 is the angle ABC).

3. Calculate distance AC: We have two sides (AB=8km, BC=6km) and the included angle (ABC=30°). Use the Cosine Rule.
   $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\text{ABC})$
   $AC^2 = 8^2 + 6^2 - 2(8)(6)\cos(30^\circ)$
   $AC^2 = 64 + 36 - 96 \times 0.866$ (approx $\cos 30^\circ$)
   $AC^2 = 100 - 83.136 = 16.864$
   $AC = \sqrt{16.864} \approx 4.107 \text{ km}$

4. Calculate bearing of C from A:

   • First, find angle BAC using the Sine Rule (we now know side AC).
     $\frac{\sin(\text{BAC})}{BC} = \frac{\sin(\text{ABC})}{AC}$
     $\frac{\sin(\text{BAC})}{6} = \frac{\sin(30^\circ)}{4.107}$
     $\sin(\text{BAC}) = \frac{6 \times 0.5}{4.107} = \frac{3}{4.107} \approx 0.7303$
     $\text{BAC} = \arcsin(0.7303) \approx 46.9^\circ$
   • The bearing of B from A is 060°. Angle BAC is about 46.9°.
   • Bearing of C from A = Bearing of B from A + Angle BAC = 060° + 46.9° = 106.9°.
   • As a three-figure bearing, this is 107°.

4. Key Takeaways

• Always draw clear diagrams, especially with North lines and bearings.
• Bearings are always measured clockwise from North and written with three figures.
• Pythagoras' theorem is only for right-angled triangles to find unknown sides.
• Use the Sine Rule when you have pairs of opposite known side/angle or two angles and one side.
• Use the Cosine Rule when you have two sides and the included angle (SAS) or all three sides (SSS).
• Remember parallel line properties (alternate, corresponding, interior angles) when working with bearings.
• Take care when calculating interior angles within a triangle using bearings.

Common Mistakes to Avoid:
- Using Pythagoras or SOH CAH TOA for non-right-angled triangles.
- Incorrectly calculating the interior angle of a triangle when given bearings.
- Forgetting to write bearings as three figures (e.g., 45° should be 045°).
- Mixing up which angle corresponds to which side in the Sine and Cosine rules.

5. Now Try It

A plane flies from Airport P to Airport Q, a distance of 100 km, on a bearing of 070°. It then flies from Airport Q to Airport R, a distance of 150 km. The bearing of R from P is 120°. Calculate the distance PR and the bearing of R from Q.

Success looks like: You should be able to draw an accurate diagram, correctly identify the interior angles of triangle PQR, use the Sine Rule to find a missing angle, and then apply the Cosine Rule to find the distance PR. Finally, you'll need to use bearings and angles to find the bearing of R from Q. (Hint: Find angle PQR first).