Mathematics Standard • Year 12 • Trigonometry • Lesson 8
Compass Radial Surveys, Skill Drill
Work one survey all the way through, in the order the method asks for: angles first, close the survey, then areas, then boundaries. Every question below uses the same field, so an error early on will show up later, which is the point.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 What two measurements does a radial survey record for each corner?
Q1.2 How do you find the angle at O between two neighbouring radial lines?
Q1.3 Which rule gives the area of one triangle, and which gives a boundary length?
Q1.4 The angles at the central point must total ____________ degrees.
2. Worked example, the wrap-around angle
The one angle in every survey that cannot be found by straight subtraction.
Problem. A survey from O has its last corner D on 305°T and its first corner A on 041°T. Find angle DOA.
Step 1, see why subtraction fails.
305° − 041° = 264°, which is a reflex angle.
Reason: the two bearings sit either side of north, so subtracting measures the long way round the circle, not the gap between the two radial lines.
Step 2, travel clockwise in two pieces.
D to north: 360° − 305° = 55°. North to A: 041°.
Reason: bearings restart at 0° at north, so the journey has to be split there.
Step 3, add and check.
angle DOA = 55° + 41° = 96°, and 264° + 96° = 360° as expected.
Reason: the wrong answer and the right one are the two angles on either side of the same pair of lines, so they must add to a full turn.
3. One survey, all the way through
A field KLMN is surveyed from a central point O. The record reads:
K · 025°T · 50 m L · 100°T · 65 m M · 186°T · 58 m N · 295°T · 72 m
Q3.1 Sketch the survey. Mark north at O and label each radial line with its bearing and length. 2 marks
Q3.2 Find all four angles at O, then check the survey closes. 3 marks
Q3.3 Find the area of triangle KOL, correct to 2 decimal places. 1 mark
Q3.4 Find the areas of the other three triangles, each to 2 decimal places, then the total area of the field to the nearest square metre. 4 marks
4. Boundaries
Same field. Give answers to 2 decimal places. 2 marks each
Q4.1 Find the length of boundary KL.
Q4.2 Find the length of boundary MN. Say what the obtuse angle at O does to the correction term.
Q4.3 Find the length of boundary NK, and explain why this one could have been done without the cosine rule at all.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, What is recorded
For each corner, the true bearing of that corner from the central point, and the distance to it. Nothing else, and in particular no boundary is ever measured.
Q1.2, The angle at O
Subtract the two bearings. For the wrap-around gap that crosses north, take 360° minus the larger bearing and add the smaller one.
Q1.3, Which rule for what
Area rule, ½ ab sin C, for the area of a triangle. Cosine rule, c² = a² + b² − 2ab cos C, for a boundary length. Both take the two radii as a and b and the angle at O as C.
Q1.4, The check
360 degrees.
Q3.2, The four angles
∠KOL = 100° − 025° = 75°, ∠LOM = 186° − 100° = 86°, ∠MON = 295° − 186° = 109° [2].
Wrap-around: ∠NOK = (360° − 295°) + 025° = 65° + 25° = 90°. Check: 75 + 86 + 109 + 90 = 360°, so every subtraction is right [1].
Q3.3, Area of KOL
½(50)(65) sin75° = 1625 × 0.965926 = 1569.63 m².
Q3.4, The other three, and the total
[LOM] = ½(65)(58) sin86° = 1885 × 0.997564 = 1880.41 m² [1]
[MON] = ½(58)(72) sin109° = 2088 × 0.945519 = 1974.24 m² [1]
[NOK] = ½(72)(50) sin90° = 1800 × 1 = 1800.00 m² [1]
Total = 1569.63 + 1880.41 + 1974.24 + 1800.00 = 7224.28, which is 7224 m² to the nearest square metre [1]. Round only here, not at each triangle.
Q4.1, Boundary KL
KL² = 50² + 65² − 2(50)(65) cos75° = 6725 − 6500(0.258819) = 6725 − 1682.32 = 5042.68 [1], so KL = 71.01 m [1].
Q4.2, Boundary MN
MN² = 58² + 72² − 2(58)(72) cos109° = 8548 − 8352(−0.325568) = 8548 + 2719.15 = 11267.15 [1], so MN = 106.15 m [1].
Because the angle at O is obtuse, cos109° is negative and the correction term is added. MN therefore comes out longer than the Pythagoras value √8548 = 92.46 m. A wide angle at O pushes the two corners further apart.
Q4.3, Boundary NK
NK² = 72² + 50² − 2(72)(50) cos90° = 7684 − 0 = 7684, so NK = 87.66 m [1].
The angle at O is exactly 90° and cos90° = 0, so the correction term vanishes and the cosine rule collapses into Pythagoras. Triangle NOK is right-angled at O, so NK² = 72² + 50² could have been written straight down [1]. This is the general fact from the lesson seen in one particular triangle: the cosine rule is Pythagoras with a correction, and at 90° there is nothing to correct.