Bearings, Scale Drawing and Constructions: past paper questions & worked solutions

**Step 1:** Place the point of the compass at the vertex *Q*. Draw an arc that intersects both arms *Q**P* and *Q**R*. Let’s call these intersection points *A* and *B*. **Step 2:** Without changing the compass width, place the compass point on *A* and draw an arc inside the angle. **Step 3:** Place the compass point on *B* and draw another arc to intersect the previous arc. Let this intersection be point *C*. **Step 4:** Draw a straight line from *Q* through *C*. This is the angle bisector.

To construct the bisector of angle \(ABC\):

1. Use centre \(B\) to draw an arc crossing \(BA\) and \(BC\). 2. Use the same compass width from the two crossing points to make two arcs inside the angle. 3. Join \(B\) to the point where these two arcs meet.

To construct the perpendicular bisector of \(AB\):

1. Open the compass to more than half the length of \(AB\). 2. Draw equal arcs from \(A\) above and below \(AB\). 3. Without changing the compass width, draw equal arcs from \(B\). 4. Join the two arc intersections with a ruler.

Use a compass radius greater than half of \(AB\). Draw matching arcs from \(A\) above and below the line, then draw matching arcs from \(B\) with the same radius. Join the two arc-intersection points.

To construct the bisector of angle \(ABC\):

1. Draw an arc centred at \(B\) cutting both arms of the angle. 2. From the two cut points, draw equal arcs inside the angle. 3. Draw the line from \(B\) through the intersection of these arcs.

The bearing of Paris from London is the reverse bearing of London from Paris.

Reverse bearings differ by \(180^\circ\): \[ 149^\circ+180^\circ=329^\circ \]

To construct the bisector of angle \(BAC\):

1. Put the compass point at \(A\) and draw an arc that cuts \(AB\) and \(AC\). 2. From the two points where the arc cuts the arms, draw equal arcs inside the angle. 3. Draw a straight line from \(A\) through the intersection of those two arcs.

To construct the perpendicular bisector of \(AB\):

1. Set the compass radius to more than half of \(AB\). 2. With centre \(A\), draw arcs above and below the line. 3. With the same radius and centre \(B\), draw arcs crossing the first pair. 4. Draw the straight line through the two arc intersections.

To construct the bisector of angle \(ABC\):

1. Put the compass point at \(B\) and draw an arc cutting \(BA\) and \(BC\). 2. From those two cut points, draw equal arcs inside the angle. 3. Draw a straight line from \(B\) through the point where the two arcs cross.

The distance from \(A\) to \(C\) is \(700\text{ m}\).

So the map distance is \[ 700\div 200=3.5\text{ cm} \]

Part (a): from \(A\), draw a bearing of \(110^\circ\) and mark \(C\) at \(3.5\text{ cm}\) from \(A\).

For part (b), \[ 200\text{ m}=20000\text{ cm} \]

So the scale is \[ 1:20000 \]

The bearing of \(B\) from \(A\) is \(054^\circ\), so the bearing of \(A\) from \(B\) is \[ 054^\circ+180^\circ=234^\circ \]

The bearing of \(C\) from \(B\) is \(132^\circ\).

So \[ \angle ABC=234^\circ-132^\circ=102^\circ \]

Use the cosine rule to find \(AC\): \[ AC^2=3.6^2+8.4^2-2(3.6)(8.4)\cos 102^\circ \]

Total distance walked: \[ 3.6+8.4+9.8028\ldots=21.8028\ldots\text{ km} \]

At \(6\text{ km/h}\), time is \[ 21.8028\ldots\div 6=3.6338\ldots\text{ hours} \]

From \(A\), ship \(C\) is on a bearing of \(120^\circ\), and \(AC=275\text{ m}\).

Resolve \(AC\): \[ \text{east component}=275\sin 120^\circ=238.16\ldots \] \[ \text{north component}=275\cos 120^\circ=-137.5 \]

Ship \(B\) is \(150\text{ m}\) north of \(A\), so from \(B\) to \(C\): \[ \text{east}=238.16\ldots,\qquad \text{north}=-137.5-150=-287.5 \]

The bearing is therefore in the south-east quadrant: \[ 180^\circ-\tan^{-1}\left(\frac{238.16\ldots}{287.5}\right)=140.36\ldots^\circ \]

Resolve each journey into east and north components.

From \(A\) to \(B\):

From \(B\) to \(C\):

So from \(A\) to \(C\):

Therefore from \(C\) to \(A\):

This direction is west of north.

Bearing of \(A\) from \(C\):