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Trigonometry - Bearings

Grade 10IGCSE

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

🔑Concepts

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Bearings are always measured from the North line.

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Bearings are always measured in a clockwise direction.

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Bearings must be written as three figures (e.g., 045∘045^\circ instead of 45∘45^\circ).

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Back Bearings: The bearing of point A from point B is the reverse of the bearing of B from A (Add/subtract 180∘180^\circ).

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Parallel lines: All North lines are parallel, so interior angles sum to 180∘180^\circ or alternate angles are equal.

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Trigonometric applications: Using SOH CAH TOA for right-angled triangles or Sine/Cosine rules for non-right-angled triangles formed by bearing paths.

📐Formulae

Back Bearing=(Forward Bearing+180∘)(mod360∘)\text{Back Bearing} = (\text{Forward Bearing} + 180^\circ) \pmod{360^\circ}

sin⁡θ=OppositeHypotenuse,cos⁡θ=AdjacentHypotenuse,tan⁡θ=OppositeAdjacent\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}, \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}

\text{Sine Rule: } \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

\text{Cosine Rule: } a^2 = b^2 + c^2 - 2bc \cos A$

\text{Area of a Triangle: } \frac{1}{2}ab \sin C$

💡Examples

Problem 1:

A ship sails 12 km East and then 9 km North. Calculate the bearing of the ship from its starting point.

Solution:

  1. Represent movement as a right-angled triangle. O=12O = 12 (East), A=9A = 9 (North).
  2. Calculate the internal angle θ\theta from the North line: tan⁡θ=129\tan \theta = \frac{12}{9}.
  3. θ=arctan⁡(1.333)≈53.1∘\theta = \arctan(1.333) \approx 53.1^\circ.
  4. Since the movement is North then East, the angle is measured clockwise from North. Bearing = 053.1∘053.1^\circ.

Explanation:

We use the tangent ratio because we have the opposite (Eastward distance) and adjacent (Northward distance) sides relative to the North line at the starting point.

Problem 2:

Point B is on a bearing of 135∘135^\circ from Point A. Find the bearing of Point A from Point B.

Solution:

  1. Given bearing 135∘135^\circ.
  2. Since 135<180135 < 180, add 180∘180^\circ.
  3. 135∘+180∘=315∘135^\circ + 180^\circ = 315^\circ.

Explanation:

To find a back bearing (the direction to return to the start), add 180∘180^\circ if the original bearing is less than 180∘180^\circ, or subtract 180∘180^\circ if it is greater.

Problem 3:

A plane flies from airport P for 200 km on a bearing of 060∘060^\circ to point Q. It then changes course and flies 150 km on a bearing of 150∘150^\circ to point R. Find the distance PR.

Solution:

  1. Draw a sketch. Angle PQRPQR is needed.
  2. North line at Q: Interior angle with P is 180−60=120∘180 - 60 = 120^\circ.
  3. Angle around Q: 360−120−150=90∘360 - 120 - 150 = 90^\circ.
  4. Using Pythagoras (PR2=PQ2+QR2PR^2 = PQ^2 + QR^2): PR=2002+1502=40000+22500=250PR = \sqrt{200^2 + 150^2} = \sqrt{40000 + 22500} = 250 km.

Explanation:

By using the properties of parallel North lines, we determined the internal angle between the two paths was 90∘90^\circ, allowing us to use the Pythagorean theorem to find the direct distance.