How to Solve Earth Geometry Questions in ECZ Mathematics

Mathematics (Grade 12 & GCE)  |  Updated March 2026  |  12 min read

Earth Geometry is a compulsory, high-scoring section in ECZ Grade 12 and GCE Mathematics Paper 2 Section B (worth 6 to 8 marks). Success requires clear mastery of latitude/longitude coordinates and distance formulas.

Senior Secondary Tip

Always take Earth radius $R = 6370\text{ km}$ or $R = 3437\text{ nm}$ (or $60\text{ nm per degree}$) exactly as specified on the front cover of your ECZ exam paper!

1. Understanding Latitudes & Longitudes

  • Latitudes ($\alpha$): Lines running East-West around the globe (Equator $0^\circ$, North $N$, South $S$).
  • Longitudes ($\theta$): Lines running North-South passing through the poles (Greenwich Meridian $0^\circ$, East $E$, West $W$).
  • Finding Angular Difference ($\theta$):
    • Same hemisphere (e.g., $30^\circ N$ and $70^\circ N$): Subtract angles ($\theta = 70 - 30 = 40^\circ$).
    • Opposite hemispheres (e.g., $20^\circ N$ and $30^\circ S$): Add angles ($\theta = 20 + 30 = 50^\circ$).

2. Distance Along a Great Circle (Longitude / Equator)

For distance along a line of longitude or along the Equator:

\[ \text{Distance in km} = \frac{\theta}{360^\circ} \times 2 \pi R \] \[ \text{Distance in nautical miles (nm)} = \theta \times 60\text{ nm} \]

3. Distance Along a Small Circle (Parallel of Latitude)

For distance along a line of latitude $\alpha^\circ N/S$ (where $r = R \cos \alpha$):

\[ \text{Distance in km} = \frac{\theta}{360^\circ} \times 2 \pi R \cos \alpha \] \[ \text{Distance in nautical miles (nm)} = \theta \times 60 \cos \alpha \]

ECZ Past Paper Style Problem:

Two towns $P(60^\circ N, 30^\circ W)$ and $Q(60^\circ N, 50^\circ E)$ lie on the surface of the Earth.

  1. Find the difference in longitude between $P$ and $Q$.
  2. Calculate the distance $PQ$ along the line of latitude in nautical miles.

Solution Part (i):

Longitudes are in opposite hemispheres ($30^\circ W$ and $50^\circ E$):

\[ \theta = 30^\circ + 50^\circ = 80^\circ \]

Solution Part (ii):

Using small circle formula with latitude $\alpha = 60^\circ N$:

\[ \text{Distance } PQ = \theta \times 60 \cos(60^\circ) \] \[ \text{Distance } PQ = 80 \times 60 \times 0.5 = 2400\text{ nautical miles} \]
Time Difference Calculation

Earth rotates $360^\circ$ in 24 hours, which equals **$15^\circ$ per hour** (or $1^\circ$ every 4 minutes). Locations further EAST are ahead in time!

Teacher Strategy

Draw a clear 3D sphere diagram on the chalkboard for every question, labeling the equator, Greenwich meridian, and given towns before substituting numbers into formulas.

Frequently Asked Questions

Q1: What is the difference between a Great Circle and a Small Circle?

A Great Circle has a radius equal to the Earth's radius R (e.g., the Equator and all lines of longitude). A Small Circle has a radius r smaller than Earth's radius (e.g., all lines of latitude except the Equator), where r = R * cos(θ).

Q2: How do you calculate distance in Nautical Miles along a Great Circle?

Distance in nautical miles (nm) along a great circle = θ * 60, where θ is the angular difference in degrees.

Q3: What is the formula for distance along a Small Circle (Parallel of Latitude)?

Distance = (θ / 360) * 2 * π * R * cos(latitude) in kilometers, OR Distance = θ * 60 * cos(latitude) in nautical miles.

Ace Your ECZ Mathematics Paper 2

Download Earth Geometry past paper questions with step-by-step worked solutions on ECZ Solutions.

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