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Trigonometry - Bearings

Grade 9IGCSE

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

🔑Concepts

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Bearings are measured from the North line, clockwise, and must always be written with three digits (e.g., 045∘045^\circ instead of 45∘45^\circ).

A bearing of 057 degrees shown clockwise from the North line at point A to point B.
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The bearing from AA to BB and the bearing from BB to AA (Back Bearing) are related by 180∘180^\circ. Since North lines are parallel, interior angles sum to 180∘180^\circ.

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To solve complex bearing problems involving non-right-angled triangles, the Sine Rule and Cosine Rule are essential.

A standard triangle with sides a, b, c and vertices A, B, C used for Sine and Cosine rules.
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Always draw a fresh North line at every point where a direction change occurs to accurately identify angles.

📐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}}

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

Area of a Triangle: 12absin⁡C\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.

Problem 4:

Town YY is 5050 km from Town XX on a bearing of 250∘250^\circ. Calculate how far South Town YY is from Town XX.

Diagram showing point Y at 250 degrees from X, forming a triangle with distance d south.

Solution:

  1. Draw the North line at XX. The bearing 250∘250^\circ is in the 3rd quadrant.
  2. The angle measured clockwise from North is 250∘250^\circ.
  3. The angle inside the right-angled triangle between the West line and the path XYXY is 250∘−270∘250^\circ - 270^\circ is not ideal; instead, use the angle from the South line: 250∘−180∘=70∘250^\circ - 180^\circ = 70^\circ.
  4. Let the southward distance be dd. In the triangle: cos⁡(70∘)=AdjacentHypotenuse=d50\cos(70^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{d}{50}
  5. d=50×cos⁡(70∘)≈17.1 kmd = 50 \times \cos(70^\circ) \approx 17.1 \text{ km}

Explanation:

We use the definition of bearings to create a right-angled triangle. By finding the angle relative to the South line (250∘−180∘=70∘250^\circ - 180^\circ = 70^\circ), we use the cosine ratio to find the vertical (South) component.

Problem 5:

A hiker walks 88 km on a bearing of 040∘040^\circ from AA to BB, and then 1212 km on a bearing of 110∘110^\circ from BB to CC. Find the distance ACAC.

Triangle ABC formed by two bearing paths, showing points A, B, and C with North lines at A and B.

Solution:

  1. At point BB, draw a North line. The interior angle between the North line at BB and the segment ABAB is 180∘−040∘=140∘180^\circ - 040^\circ = 140^\circ (using parallel lines/consecutive interior angles).
  2. The angle ABCABC is the difference between the full rotation (360∘360^\circ) and the sum of the back-bearing angle and the new bearing: Angle ∠ABC=180∘−(110∘−40∘)=110∘\angle ABC = 180^\circ - (110^\circ - 40^\circ) = 110^\circ. Alternatively, using North lines: The angle between South and BCBC is 110∘−180∘110^\circ - 180^\circ (not useful) or simply realize the angle inside the triangle at BB is 180∘−110∘+40∘=110∘180^\circ - 110^\circ + 40^\circ = 110^\circ.
  3. Use the Cosine Rule: AC2=82+122−2(8)(12)cos⁡(110∘)AC^2 = 8^2 + 12^2 - 2(8)(12)\cos(110^\circ)
  4. AC2=64+144−192(−0.342)AC^2 = 64 + 144 - 192(-0.342)
  5. AC2=208+65.66=273.66AC^2 = 208 + 65.66 = 273.66
  6. AC=273.66≈16.5 kmAC = \sqrt{273.66} \approx 16.5 \text{ km}

Explanation:

To find the distance between two points after a turn, we determine the interior angle of the triangle formed by the paths. Here, we calculate ∠ABC\angle ABC using parallel North lines and then apply the Cosine Rule.