Mapping a country with one measured line
Tonight's idea
You cannot stretch a tape measure across a mountain range. You can measure angles across one, with a telescope on a turntable. Triangulation turns one measured length and many angles into a map.
Pick two points \(A\) and \(B\) on flat ground and measure the distance between them with great care. This is the baseline. From \(A\), sight a distant hilltop \(C\) and measure the angle between \(AB\) and \(AC\). Do the same at \(B\). The third angle is whatever is left of 180°.
Now you know all three angles and one side. The sine rule (see Triangles without a right angle) gives the other two sides:
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\), where each side is named after the angle opposite it.
The new sides are now known lengths. Use one of them as the baseline of the next triangle and you need not measure a length again. Before satellites, whole countries were mapped this way: a net of triangles grown from a few measured baselines.
Watch one: a 5 km baseline
The baseline is \(AB = c = 5\) km. The measured angles are \(A = 62°\) and \(B = 75°\), so \(C = 180° - 62° - 75° = 43°\). The known pair is \(c\) and \(C\).
\[AC = b = \frac{5 \sin 75°}{\sin 43°} = 7.08 \text{ km}, \qquad BC = a = \frac{5 \sin 62°}{\sin 43°} = 6.47 \text{ km}.\]
Check: the largest angle, 75°, faces the longest side, 7.08 km, and the smallest angle faces the shortest side.
Enter two angles and a side (ASA) and watch the working: mathdetours.com/demos/trianglesolver.html.
Your turn
- A baseline \(AB\) is 8 km long. The angle at \(A\) is 50° and the angle at \(B\) is 60°. Find the angle at \(C\), then \(AC\) and \(BC\).
- Two lighthouses \(A\) and \(B\) stand 12 km apart on a straight coast. A ship \(C\) is sighted at 40° to the coast from \(A\) and at 65° to the coast from \(B\). Find \(AC\), then the ship's distance from the coast.
- Carry on from the example, with \(BC = 6.47\) km as the new baseline. A tower \(D\) is sighted, and the angles of triangle \(BCD\) are 64° at \(B\) and 56° at \(C\). Find \(CD\) and \(BD\).
- A baseline is 5 km, with \(A = 88°\) and \(B = 89°\). Find \(AC\). Now suppose the angle at \(B\) was really 89.5°. Find \(AC\) again. What shape of triangle should a surveyor choose, and why?
Stretch: one mistake, everywhere
Every length in the net is the baseline times a ratio of sines. Years later, the baseline is found to have been measured 0.1% too long. How far out is a distance of 400 km computed across the net? What happens to the angles on the map?
Answers
- \(C = 180° - 50° - 60° = 70°\). \(AC = \dfrac{8 \sin 60°}{\sin 70°} = 7.37\) km (opposite \(B\)). \(BC = \dfrac{8 \sin 50°}{\sin 70°} = 6.52\) km (opposite \(A\)).
- \(C = 180° - 40° - 65° = 75°\). \(AC = \dfrac{12 \sin 65°}{\sin 75°} = 11.26\) km. Drop a perpendicular from \(C\) to the coast: its length is \(AC \sin 40° = 7.24\) km.
- \(D = 180° - 64° - 56° = 60°\). \(CD\) is opposite \(B\): \(CD = \dfrac{6.47 \sin 64°}{\sin 60°} = 6.71\) km. \(BD\) is opposite \(C\): \(BD = \dfrac{6.47 \sin 56°}{\sin 60°} = 6.19\) km.
- \(C = 3°\), so \(AC = \dfrac{5 \sin 89°}{\sin 3°} = 95.5\) km. With \(B = 89.5°\), \(C = 2.5°\) and \(AC = \dfrac{5 \sin 89.5°}{\sin 2.5°} = 114.6\) km. Half a degree moved the answer by 20%. The trouble is the tiny angle at \(C\): its sine is on the bottom of the fraction, and a small change in a small number is a large change in proportion. Choose triangles that are close to equilateral. With all three angles 60°, the same half-degree error changes \(AC\) from 5.00 km to 5.05 km, about 1%.
Stretch. Every length is multiplied by the same factor, 1.001. A 400 km distance is out by \(0.001 \times 400 = 0.4\) km, or 400 m. The angles do not change at all: scaling a figure keeps its angles. The map has the right shape and the wrong size, which is why the baseline is the one measurement worth extreme care.