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Trigonometry - Bearings

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Three-Figure Notation: Bearings must always be written with three digits (e.g., 045° instead of 45°).

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Measured from North: Always start measuring from the North line (000°).

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Clockwise Direction: Bearings are always measured in a clockwise direction from North.

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Back Bearings: To find the bearing of A from B when given the bearing of B from A, add or subtract 180°. If the bearing is < 180°, add 180°; if it is > 180°, subtract 180°.

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Parallel Lines: North lines are always parallel. Use properties like interior angles (sum to 180°) and alternate angles (Z-angles) to find missing angles.

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Trigonometric Application: Most IGCSE bearing problems require the Sine Rule or Cosine Rule to find missing distances or angles between three points.

📐Formulae

Back Bearing=θ±180∘\text{Back Bearing} = \theta \pm 180^\circ

Sine Rule:asin⁡A=bsin⁡B=csin⁡C\text{Sine Rule}: \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Cosine Rule (Side):a2=b2+c2−2bccos⁡A\text{Cosine Rule (Side)}: a^2 = b^2 + c^2 - 2bc \cos A

Cosine Rule (Angle):cos⁡A=b2+c2−a22bc\text{Cosine Rule (Angle)}: \cos A = \frac{b^2 + c^2 - a^2}{2bc}

SOH CAH TOA (for right-angled triangles)\text{SOH CAH TOA (for right-angled triangles)}

💡Examples

Problem 1:

A ship sails 12 km on a bearing of 070° from port P to point A. It then sails 15 km on a bearing of 150° from A to point B. Calculate the distance PB.

Solution:

  1. Angle at A: The interior angle between the North line at P and the North line at A is 180−70=110∘180 - 70 = 110^\circ.
  2. The angle around point A includes the bearing of B (150°) and the interior angle. To find the internal angle ∠PAB\angle PAB: 360∘−110∘−(360−150)=…360^\circ - 110^\circ - (360 - 150) = \dots or more simply: Angle between South at A and AB is 150−180=−30150 - 180 = -30 (invalid), use: 180−(150−70)=100∘180 - (150-70) = 100^\circ or visualize: ∠PAB=(180−70)+150\angle PAB = (180 - 70) + 150 is not right. Correct logic: Angle at A relative to North is 70° (alternate). So angle inside triangle is 180−70+150180 - 70 + 150 is wrong. Correct: (180−70)=110(180 - 70) = 110. Bearing of B from A is 150. So ∠PAB=180−(150−70)=100∘\angle PAB = 180 - (150 - 70) = 100^\circ.
  3. Use Cosine Rule: PB2=122+152−2(12)(15)cos⁡(100∘)PB^2 = 12^2 + 15^2 - 2(12)(15)\cos(100^\circ).
  4. PB2=144+225−360(−0.1736)=369+62.5=431.5PB^2 = 144 + 225 - 360(-0.1736) = 369 + 62.5 = 431.5.
  5. PB=431.5≈20.8PB = \sqrt{431.5} \approx 20.8 km.

Explanation:

To solve complex bearings, always draw the North lines at every point. Use the 'Z-rule' (alternate angles) or interior angles to find the internal angle of the triangle formed, then apply the Cosine Rule for the unknown side.

Problem 2:

The bearing of a lighthouse L from a boat B is 240°. What is the bearing of the boat from the lighthouse?

Solution:

  1. Given Bearing B→L=240∘B \to L = 240^\circ.
  2. Since 240∘>180∘240^\circ > 180^\circ, subtract 180∘180^\circ.
  3. 240∘−180∘=060∘240^\circ - 180^\circ = 060^\circ.

Explanation:

This is a back-bearing problem. Since the North lines are parallel, the angles are related by 180 degrees. If you are looking at someone on a bearing of 240°, they are looking back at you on a bearing of 060°.