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Geometry and Measurement - Bearings

Grade 8IB

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

🔑Concepts

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Bearings are always measured starting from North, moving in a clockwise direction, and are written as three-digit figures (e.g., 045∘045^\circ instead of 45∘45^\circ).

A diagram showing a bearing of 060 degrees measured clockwise from the North line at point A to point B.
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Parallel line rules are essential for bearing problems. The North lines at two different points are always parallel to each other. Consequently, interior angles between these lines sum to 180∘180^\circ, and alternate angles are equal.

Two parallel North lines at points A and B connected by a line segment, illustrating interior angles.
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The back bearing is the bearing from the destination back to the starting point. It can be found by adding or subtracting 180∘180^\circ from the original bearing, ensuring the result is between 000∘000^\circ and 360∘360^\circ.

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Trigonometry (SOH CAH TOA) can be used to calculate bearings when horizontal and vertical distances between two points are known. Always draw a right-angled triangle relative to the North line.

📐Formulae

Back Bearing=θ+180∘ (if θ<180∘)\text{Back Bearing} = \theta + 180^\circ \text{ (if } \theta < 180^\circ \text{)}

Back Bearing=θ−180∘ (if θ≥180∘)\text{Back Bearing} = \theta - 180^\circ \text{ (if } \theta \ge 180^\circ \text{)}

Interior Angles Sum=180∘\text{Interior Angles Sum} = 180^\circ

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

💡Examples

Problem 1:

The bearing of a lighthouse LL from a ship SS is 075∘075^\circ. Find the bearing of the ship SS from the lighthouse LL.

Solution:

Bearing=75∘+180∘=255∘\text{Bearing} = 75^\circ + 180^\circ = 255^\circ

Explanation:

To find the back bearing from LL to SS, we add 180∘180^\circ to the original bearing because 075∘075^\circ is less than 180∘180^\circ. This follows the rule for reverse directions on parallel North lines.

Problem 2:

A hiker walks 10 km10\text{ km} due North from point AA to point BB, and then 10 km10\text{ km} due East to point CC. Calculate the bearing of CC from AA.

Solution:

tan⁡(θ)=1010=1\tan(\theta) = \frac{10}{10} = 1 θ=arctan⁡(1)=45∘\theta = \arctan(1) = 45^\circ Bearing=045∘\text{Bearing} = 045^\circ

Explanation:

The movement forms a right-angled triangle with the North line. The angle θ\theta from the North line at AA is found using the tangent ratio (Opposite/Adjacent). Since the angle is 45∘45^\circ, we write it as a three-figure bearing: 045∘045^\circ.

Problem 3:

If the bearing of YY from XX is 310∘310^\circ, calculate the bearing of XX from YY.

Solution:

310−180130\begin{array}{r} 310 \\ - 180 \\ \hline 130 \end{array} Bearing=130∘\text{Bearing} = 130^\circ

Explanation:

Since the original bearing 310∘310^\circ is greater than 180∘180^\circ, we subtract 180∘180^\circ to find the back bearing. 310∘−180∘=130∘310^\circ - 180^\circ = 130^\circ.

Problem 4:

A plane flies from Airport PP to Airport QQ on a bearing of 125∘125^\circ. Calculate the bearing of PP from QQ.

Diagram showing point P with a North line and a path to Q at a bearing of 125 degrees.

Solution:

The bearing of QQ from PP is θ=125∘\theta = 125^\circ. Since 125∘<180∘125^\circ < 180^\circ, we calculate the back bearing as: Back Bearing=125∘+180∘=305∘\text{Back Bearing} = 125^\circ + 180^\circ = 305^\circ Therefore, the bearing of PP from QQ is 305∘305^\circ.

Explanation:

Because the North lines at PP and QQ are parallel, the angle from QQ back to PP is found by considering the interior angles (which add to 180∘180^\circ) and the full circle at QQ. Adding 180∘180^\circ effectively reverses the direction.

Problem 5:

A ship sails 40 km40\text{ km} due East from point AA and then 30 km30\text{ km} due South to point BB. Calculate the bearing of BB from AA to the nearest degree.

A right-angled triangle showing 40 km East and 30 km South from point A to point B.

Solution:

  1. Let α\alpha be the angle between the North line and the path ABAB.
  2. Since the ship goes East and then South, it forms a right-angled triangle.
  3. The angle from the East direction to BB can be found using: tan⁡(θ)=OppositeAdjacent=3040\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{30}{40} θ=arctan⁡(0.75)≈36.87∘\theta = \arctan(0.75) \approx 36.87^\circ
  4. The bearing is measured from North. East is 090∘090^\circ. Bearing=90∘+36.87∘=126.87∘\text{Bearing} = 90^\circ + 36.87^\circ = 126.87^\circ To the nearest degree, the bearing is 127∘127^\circ.

Explanation:

We use the tangent ratio in the right-angled triangle formed by the East and South displacements. Since bearings start from North (000∘000^\circ), we add the calculated angle to 90∘90^\circ (East) to get the final clockwise bearing.