Mathematics
Mathematics items on the Connecticut exam reward clean geometry and disciplined sign conventions rather than exotic theory. Expect bearing–bearing and distance–distance intersections, inverses, circular curves, coordinate areas, unit conversions and error propagation.
Core tools
- Inverse: ΔN = N2 − N1, ΔE = E2 − E1; distance = √(ΔN² + ΔE²); azimuth = atan2(ΔE, ΔN), converted to a quadrant bearing.
- Forward: N = N1 + d·cos(bearing), E = E1 + d·sin(bearing), with bearings measured clockwise from north.
- Circular curves: T = R·tan(Δ/2), L = R·Δ in radians, LC = 2R·sin(Δ/2), M = R(1 − cos(Δ/2)), E = R(sec(Δ/2) − 1).
- Area by coordinates: A = ½ |Σ(Eᵢ·Nᵢ₊₁ − Eᵢ₊₁·Nᵢ)|.
- Units worth memorizing: 1 ft = 0.3048 m exactly, 1 rod = 16.5 ft, 1 chain = 66 ft, 1 acre = 43,560 ft².
Worked example (bearing–bearing intersection)
Point P lies on a line from A(1000.00, 1000.00) bearing N 30° E and on a line from B(1300.00, 1000.00) bearing N 60° W. Direction vectors: from A (sin 30, cos 30) = (0.50000, 0.86603); from B (−sin 60, cos 60) = (−0.86603, 0.50000). A + t(0.50000, 0.86603) = B + s(−0.86603, 0.50000). From northings: 0.86603t = 0.50000s, so s = 1.73205t. In eastings: 1000 + 0.500t = 1300 − 0.86603(1.73205t) = 1300 − 1.500t, so 2t = 300 and t = 150.00, s = 259.81. P = (1000 + 75.00, 1000 + 129.90) = (1075.00, 1129.90). Check from B: (1300 − 225.00, 1000 + 129.91) confirms it.
Traps
- Swapping sine and cosine by measuring the bearing from east instead of north.
- Entering 30°30′ as 30.30 decimal degrees.
- Confusing arc length with chord length when computing curve stationing.
- Dropping quadrant signs on latitudes and departures.
- Reporting a bearing greater than 90° instead of converting to a correct quadrant.