Understand the relationship between distance, angular distance (degrees and minutes) and time, using latitude and longitude to locate points on Earth's surface, and the definition of a nautical mile as one minute of arc along a great circle

Part (a) is worth one mark for recognising the points are in opposite hemispheres so the latitudes add: $18^\circ + 27^\circ = 45^\circ$. A marker deducts here for $27 - 18 = 9^\circ$, the classic 'subtract across the equator' error. Part (b) carries the rest: one mark for converting to nautical miles via $45 \times 60 = 2700$ NM (or $45^\circ = 2700'$ and $1' = 1$ NM), and one mark for $2700 \times 1.852 = 5000.4 \approx 5000$ km. Method marks survive an arithmetic slip provided the conversions $1^\circ = 60$ NM and $1$ NM $= 1.852$ km are shown, but an unlabelled or wrong angle in (a) loses the follow-through structure.

Part (a) earns one mark: both latitudes are $33^\circ$ S, so the difference in latitude is $0^\circ$. Part (b) earns one mark: both longitudes are east, so subtract, $138^\circ - 78^\circ = 60^\circ$. Part (c) earns one mark for the reasoning that the rule $1' = 1$ nautical mile only holds along a great circle; a parallel of latitude (other than the equator) is a small circle with a smaller radius, so $60$ NM per degree overstates the true distance. Markers accept any clear statement that the parallel is not a great circle; a bare 'because it is east-west' without the great-circle idea does not earn the mark.

Part (a) is one mark for $(22^\circ \text{ N}, \ 145^\circ \text{ E})$ with both direction labels present. Part (b) is one mark: both latitudes north, so subtract, $58^\circ - 22^\circ = 36^\circ$. Part (c) is one mark for $36 \times 60 = 2160$ nautical miles (using $1^\circ = 60$ NM). Part (d) is two marks: $2160 \times 1.852 = 4000.32$, rounding to $4000$ km, with one mark for the correct multiplication and one for the rounding and unit. A marker awards follow-through from a wrong (b): if a candidate writes $58 + 22 = 80^\circ$ they lose (b) but can still earn the conversion marks in (c) and (d) applied to their angle.

Part (a) - write latitude first, then longitude. Latitude is the angle north or south of the equator, longitude is the angle east or west of the Greenwich meridian. $P$ is $35^\circ$ south and $150^\circ$ east, so

Part (b) - on the equator the latitude is $0^\circ$. $Q$ shares the meridian $150^\circ$ E but sits on the equator, so its latitude is $0^\circ$:

(Always give the N/S label on the latitude and the E/W label on the longitude. A bare $35^\circ$ does not say which way from the equator.)

Same hemisphere, so subtract. Both latitudes are south of the equator, so the angular gap between the two parallels is the difference of the two angles:

So the difference in latitude is $23^\circ$. (If one city were north and one south you would add the two angles, because the equator sits between them.)

Opposite hemispheres, so add. One point is north of the equator and the other is south, so the equator lies between them and the angles add:

So the difference in latitude is $13^\circ$. (A quick check: the ship crosses the equator, so the total angle must be more than either single latitude, and $13^\circ$ is.)

Part (a) - same hemisphere (both east), so subtract. The difference in longitude is the gap between the two meridians:

So the difference in longitude is $35^\circ$.

Part (b) - convert degrees to minutes. One degree is $60$ minutes of arc, so

So the difference is $2100'$. (Minutes of arc, written with a prime $'$, are not minutes of time. The conversion $1^\circ = 60'$ is what links a longitude difference to nautical miles later on.)

Part (a) - same hemisphere, so subtract. The angular distance along the meridian is the difference in latitude:

So the angular distance is $30^\circ$.

Part (b) - one minute of arc is one nautical mile. Convert the angle to minutes, and each minute is $1$ nautical mile:

A faster route is $60$ nautical miles per degree, so $30 \times 60 = 1800$ NM.

Part (c) - multiply by $1.852$. Each nautical mile is $1.852$ km, so

So $A$ and $B$ are about $3334$ km apart. (Using the rougher rule $1^\circ \approx 111$ km along a meridian gives $30 \times 111 = 3330$ km, which agrees to the nearest few kilometres.)

Part (a) - coordinates, latitude then longitude. Cairns is $17^\circ$ south and on the meridian $147^\circ$ east:

Part (b) - same meridian, same hemisphere, so subtract. Both latitudes are south, so the angular distance is the difference of the two angles:

So the angular distance is $26^\circ$.

Part (c) - convert to nautical miles using $1' = 1$ NM. Turn the angle into minutes, then read each minute as a nautical mile:

Part (d) - convert nautical miles to kilometres. Multiply by $1.852$:

So Cairns and Hobart are about $2889$ km apart along the meridian. (Marks here are for the subtraction in (b), the $\times 60$ to nautical miles in (c), and the $\times 1.852$ in (d); a wrong "add instead of subtract" in (b) loses the chain.)

Part (a) - both points share the meridian $130^\circ$ E. Longitude measures the meridian a point sits on. $X$ and $Y$ are both on the meridian $130^\circ$ E, so there is no change in longitude:

Sailing "due south" keeps the longitude fixed, which is exactly why the path lies along one meridian.

Part (b) - opposite hemispheres, so add the latitudes. $X$ is north of the equator and $Y$ is south, so the equator lies between them:

So the angular distance sailed is $24^\circ$.

Part (c) - convert the angle to a distance. Each degree is $60$ nautical miles ($1' = 1$ NM), so

Then convert to kilometres by multiplying by $1.852$:

So the yacht sails $1440$ nautical miles, about $2667$ km. (The trap is part (b): because the yacht crosses the equator you add $8$ and $16$ to get $24^\circ$, you do not subtract to get $8^\circ$.)