Maps and Three Figure Bearings for the TMUA

Scale Drawings and Maps

Scale drawings are essential for representing large objects or distances on paper while maintaining geometric similarity. For the TMUA, you must be comfortable with the ratio notation $1 : n$. This notation signifies that one unit of length on the drawing or map represents $n$ units of the same length in the real world.

Unit Conversion and Scaling

When working with maps, scales are often given without units. A scale of $1 : 50,000$ means that $1$ cm on the map represents $50,000$ cm on the ground. To interpret these effectively, you must be proficient in converting metric units: $100$ cm is $1$ m, and $1,000$ m is $1$ km.

Worked Example: Calculating Actual Distance

A map has a scale of $1 : 25,000$. The distance between two points on the map is $8.4$ cm. Calculate the actual distance in kilometres.

1. Multiply the map distance by the scale factor: $8.4 \times 25,000 = 210,000$ cm.
2. Convert centimetres to metres: $210,000 \div 100 = 2,100$ m.
3. Convert metres to kilometres: $2,100 \div 1,000 = 2.1$ km.

Finding Drawing Lengths

Conversely, to find the drawing length from an actual distance, you divide the actual distance by the scale factor, ensuring the units match first.

Worked Example: Calculating Map Length

A building is $15$ m long. How long would it be on a plan with a scale of $1 : 200$, in centimetres?

1. Convert the actual distance to the desired units: $15$ m = $1,500$ cm.
2. Divide by the scale factor: $1,500 \div 200 = 7.5$ cm.

Three Figure Bearings

Bearings provide a standardised way to describe the direction of one point from another. In the TMUA, you must strictly follow three specific rules for bearings:

1. They are always measured from the North line.
2. They are always measured in a clockwise direction.
3. they are always written as three figures. For example, an angle of $7^{\circ}$ must be written as $007^{\circ}$, and $60^{\circ}$ must be written as $060^{\circ}$.

Reverse Bearings

If the bearing of point $B$ from point $A$ is known, you can calculate the bearing of $A$ from $B$, often called the back bearing or reverse bearing. Because North lines at both points are parallel, we can use the geometric properties of parallel lines.

Consider the bearing of $B$ from $A$ to be $\theta$. The North line at $A$ and the North line at $B$ are parallel. The line $AB$ acts as a transversal. The interior angles between parallel lines sum to $180^{\circ}$. Therefore, if the bearing is less than $180^{\circ}$, you add $180^{\circ}$ to find the reverse bearing. If it is greater than $180^{\circ}$, you subtract $180^{\circ}$.

Worked Example: Reverse Bearings

The bearing of a lighthouse $L$ from a ship $S$ is $115^{\circ}$. Find the bearing of the ship from the lighthouse.

1. Since $115^{\circ}$ is less than $180^{\circ}$, add $180^{\circ}$.
2. $115 + 180 = 295^{\circ}$.
3. The ship is on a bearing of $295^{\circ}$ from the lighthouse.

Combining Bearings with Trigonometry

TMUA questions frequently combine bearings with the Sine and Cosine rules. To solve these, always draw a clear diagram with North lines at every vertex.

Worked Example: Bearing Problem

A hiker walks $5$ km on a bearing of $060^{\circ}$ from $P$ to $Q$, and then $8$ km on a bearing of $150^{\circ}$ from $Q$ to $R$. Find the distance $PR$.

1. At point $Q$, the North line is parallel to the North line at $P$. The interior angle property tells us the angle between $PQ$ and the North line at $Q$ (measured southwards) is $180 - 60 = 120^{\circ}$.
2. The bearing of $R$ from $Q$ is $150^{\circ}$. The angle $PQR$ is the difference between $150^{\circ}$ and the bearing from $Q$ to $P$. Alternatively, note that the bearing of $P$ from $Q$ is $180 + 60 = 240^{\circ}$. The angle $PQR = 240 - 150 = 90^{\circ}$.
3. Since $PQR$ is $90^{\circ}$, we use Pythagoras: $PR^2 = 5^2 + 8^2 = 25 + 64 = 89$.
4. $PR = \sqrt{89} \approx 9.43$ km.