Geometry, Trig & Calculus

Trigonometry: Ratios, Rules, and Bearings

SSS 1 · SSS 2 · SSS 3

✓By the end of this lesson students will be able to define and apply the basic trigonometric ratios (sine, cosine, tangent) to solve problems involving right-angled triangles.
✓By the end of this lesson students will be able to state and apply the Sine Rule and Cosine Rule to solve problems involving non-right-angled triangles.
✓By the end of this lesson students will be able to interpret and draw diagrams involving bearings (three-figure and compass bearings).
✓By the end of this lesson students will be able to solve practical problems involving heights, distances, and bearings using trigonometric principles.

Key concepts

Trigonometric Ratios (SOH CAH TOA)

Trigonometric ratios relate the angles and side lengths of right-angled triangles. For a given acute angle (θ) in a right-angled triangle:\n\n* The side opposite the angle θ is called the 'opposite' side.\n* The side adjacent to the angle θ (not the hypotenuse) is called the 'adjacent' side.\n* The side opposite the right angle is the longest side and is called the 'hypotenuse'.\n\nThe three primary ratios are:\n\n* **Sine (sin θ)**: Ratio of the length of the opposite side to the length of the hypotenuse.\n* **Cosine (cos θ)**: Ratio of the length of the adjacent side to the length of the hypotenuse.\n* **Tangent (tan θ)**: Ratio of the length of the opposite side to the length of the adjacent side.\n\nA common mnemonic to remember these ratios is SOH CAH TOA.

sin θ = Opposite / Hypotenuse\ncos θ = Adjacent / Hypotenuse\ntan θ = Opposite / Adjacent

The Sine Rule

The Sine Rule is used to solve non-right-angled triangles when you are given:\n\n* Two angles and one side (AAS or ASA).\n* Two sides and a non-included angle (SSA - this can sometimes lead to ambiguous cases).\n\nFor any triangle ABC with sides a, b, c opposite to angles A, B, C respectively, the Sine Rule states that the ratio of the length of a side to the sine of its opposite angle is constant.

a / sin A = b / sin B = c / sin C\n\nAlternatively, it can be written as:\nsin A / a = sin B / b = sin C / c

The Cosine Rule

The Cosine Rule is used to solve non-right-angled triangles when you are given:\n\n* Two sides and the included angle (SAS).\n* All three sides (SSS).\n\nFor any triangle ABC with sides a, b, c opposite to angles A, B, C respectively, the Cosine Rule relates the square of one side to the squares of the other two sides and the cosine of the included angle.

a² = b² + c² - 2bc cos A\nb² = a² + c² - 2ac cos B\nc² = a² + b² - 2ab cos C\n\nTo find an angle, the rule can be rearranged:\ncos A = (b² + c² - a²) / 2bc\ncos B = (a² + c² - b²) / 2ac\ncos C = (a² + b² - c²) / 2ab

Bearings

Bearings are used to describe the direction of one point relative to another. They are measured clockwise from the North direction.\n\nThere are two main types of bearings:\n\n1. **Three-figure bearings**: These are always expressed with three digits, from 000° to 360°. The angle is measured clockwise from the North line. For example, 045° (North-East), 180° (South), 270° (West).\n2. **Compass bearings (Cardinal Point Bearings)**: These are expressed using the cardinal points (N, S, E, W). The angle is measured from either North or South towards East or West. For example, N45°E (North 45 degrees East), S30°W (South 30 degrees West). The angle is always acute (between 0° and 90°).\n\nWhen solving problems involving bearings, it is crucial to draw accurate diagrams, indicating the North lines at each relevant point and using parallel lines properties (alternate angles, corresponding angles, interior angles) to find unknown angles within the triangles formed.

Key facts to remember

1SOH CAH TOA helps remember the basic trigonometric ratios for right-angled triangles.
2The Sine Rule (a/sin A = b/sin B = c/sin C) is used when you have AAS, ASA, or SSA (ambiguous case).
3The Cosine Rule (a² = b² + c² - 2bc cos A) is used when you have SAS or SSS.
4Bearings are measured clockwise from the North line.
5Three-figure bearings are always written with three digits (e.g., 045°, 180°, 270°).
6Compass bearings use N/S, an acute angle, and E/W (e.g., N30°E, S60°W).
7When solving bearing problems, always draw North lines at each point and use properties of parallel lines to find internal angles of triangles.

Worked examples

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, correct to one decimal place, the angle the ladder makes with the ground.

ILet the angle the ladder makes with the ground be θ.

IIDraw a right-angled triangle. The ladder is the hypotenuse (10 m), the distance from the wall is the adjacent side (6 m).

IIIIdentify the appropriate trigonometric ratio: We have the adjacent side and the hypotenuse, so we use cosine.

IVcos θ = Adjacent / Hypotenuse

Vcos θ = 6 / 10

VIcos θ = 0.6

VIIθ = cos⁻¹(0.6)

VIIIθ ≈ 53.1301°

9Round to one decimal place.

Answer

The angle the ladder makes with the ground is 53.1°.

Always draw a clear diagram to visualise the problem and label the sides correctly.

Example 2

In triangle PQR, PQ = 8 cm, QR = 12 cm and angle PQR = 70°. Calculate, correct to one decimal place, the length of PR.

IDraw triangle PQR and label the given sides and angle.

IIIdentify the given information: Two sides (PQ and QR) and the included angle (∠PQR) are given. This indicates the use of the Cosine Rule.

IIILet PR = q, PQ = r = 8 cm, QR = p = 12 cm, and ∠PQR = Q = 70°.

IVApply the Cosine Rule: q² = p² + r² - 2pr cos Q

Vq² = 12² + 8² - 2(12)(8) cos 70°

VIq² = 144 + 64 - 192 cos 70°

VIIq² = 208 - 192(0.3420) (using cos 70° ≈ 0.3420)

VIIIq² = 208 - 65.664

9q² = 142.336

10q = √142.336

11q ≈ 11.9304 cm

12Round to one decimal place.

Answer

The length of PR is 11.9 cm.

Ensure your calculator is in 'degree' mode when calculating trigonometric values.

Example 3

A ship sails from Port A on a bearing of 060° to Port B, a distance of 50 km. It then sails from Port B on a bearing of 150° to Port C, a distance of 70 km. Calculate, correct to the nearest whole number, the distance of Port C from Port A.

IDraw a diagram. Start at A, draw a North line. Draw AB at 060° for 50 km.

IIAt B, draw another North line (parallel to the first). Draw BC at 150° for 70 km.

IIIConnect A to C to form triangle ABC.

IVFind the interior angle at B (∠ABC).

VThe bearing from A to B is 060°. The angle from North at B back to A (alternate angle) is 060°.

VIThe bearing from B to C is 150°.

VII∠ABC = (Angle from North at B to line BA) + (Angle from line BA to line BC). This is incorrect. Instead, use the angle from North at B to line BA (alternate angle to 060°) and the bearing of BC.

VIIIAngle North at B to BA = 060° (alternate to bearing 060° from A).

9Angle North at B to BC = 150° (given bearing).

10∠ABC = 180° - (Angle North at B to BA) - (Angle from South at B to BC). This is also not direct.

11Let's use the angle from the North line at B to the line BA, which is 60° (alternate interior angle to the 60° bearing from A).

12The bearing of BC is 150°. This means the angle from the North line at B, clockwise to BC, is 150°.

13Therefore, the angle from the line BA to the North line at B is 60°.

14The angle from the North line at B to the line BC is 150°.

15So, ∠ABC = 180° - (angle from North at B to BA) - (angle from South at B to BC). This is getting complicated. Let's simplify.

16Angle from North at B to BA is 60° (alternate to 060°).

17Angle from North at B to BC is 150°.

18The angle between BA and the North line (pointing 'up' from B) is 60°. The angle between the North line (pointing 'up' from B) and BC is 150°.

19So, ∠ABC = 180° - 60° - (180° - 150°) = 180° - 60° - 30° = 90°. This is incorrect. Let's re-evaluate.

20Draw North line at A. Angle NAB = 60°. Draw AB = 50km.

21Draw North line at B. Angle N'BC = 150°. Draw BC = 70km.

22Angle ABN' (alternate to NAB) = 60°.

23Angle N'BC = 150°.

24Therefore, ∠ABC = N'BC - ABN' = 150° - 60° = 90°.

25Now we have triangle ABC with AB = 50 km, BC = 70 km, and ∠ABC = 90°.

26Since it's a right-angled triangle, we can use Pythagoras theorem or the Cosine Rule.

27Using Cosine Rule (for practice, though Pythagoras is simpler here): AC² = AB² + BC² - 2(AB)(BC) cos(∠ABC)

28AC² = 50² + 70² - 2(50)(70) cos 90°

29Since cos 90° = 0, the term -2(50)(70)cos 90° becomes 0.

30AC² = 2500 + 4900

31AC² = 7400

32AC = √7400

33AC ≈ 86.023 km

34Round to the nearest whole number.

Answer

The distance of Port C from Port A is 86 km.

Careful drawing of diagrams and correct identification of angles using parallel line properties are crucial for bearing problems.

Common mistakes

✗Confusing opposite and adjacent sides, especially when the angle of interest changes.
✗Using the Sine Rule or Cosine Rule for a right-angled triangle when basic SOH CAH TOA or Pythagoras' theorem would be simpler.
✗Incorrectly identifying the included angle for the Cosine Rule (SAS).
✗Errors in calculating angles in bearing problems, particularly when using alternate or corresponding angles.
✗Forgetting to put the calculator in 'degree' mode, leading to incorrect trigonometric values.

Exam tips

★Always draw a clear, well-labelled diagram for every problem, especially those involving bearings or non-right-angled triangles.
★Identify the type of triangle (right-angled or non-right-angled) and the given information (sides, angles) to choose the correct trigonometric tool (SOH CAH TOA, Sine Rule, Cosine Rule).
★Show all your working steps clearly. Marks are often awarded for correct methods even if the final answer has a minor calculation error.
★Practice converting between three-figure bearings and compass bearings, and drawing accurate bearing diagrams to avoid errors in angle calculations.

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