Mathematics Standard • Year 12 • Trigonometry • Lesson 8
Radial Survey Reasoning
Diagnose a survey that does not close, work backwards from an area to an angle, and finish with a problem that needs bearings, a non-right-angled triangle and an angle of elevation together. This is the whole focus area in one worksheet.
1. The survey that does not close
4 marks
The record. A · 035°T, B · 120°T, C · 205°T, D · 300°T.
A student's angles at O:
∠AOB = 120 − 35 = 85°
∠BOC = 205 − 120 = 85°
∠COD = 300 − 205 = 95°
∠DOA = 300 − 35 = 265°
Total = 530°
Q1.1 Identify the error, correct it, and show the survey then closes. Then say precisely what the 360° check does and does not prove. 4 marks
2. Working backwards from an area
4 marks
Q2.1 One triangle of a four-corner radial survey has radii of 64 m and 80 m, and an area of 2194.35 m².
(a) Find the two possible sizes for its angle at O. (2 marks)
(b) The other three angles at O are 85°, 92° and 124°. Say which of your two answers is correct and how you know. (2 marks)
3. Bearings, triangle and elevation together
A survey is taken from O. Corner P is 96 m away on a bearing of 018°T and corner R is 110 m away on a bearing of 168°T. A communications tower stands at R. 7 marks
Q3.1 (a) Find the angle POR. (1 mark)
(b) Find the distance PR, correct to 2 decimal places. (2 marks)
(c) From O, the angle of elevation of the top of the tower at R is 6°. Find the height of the tower, correct to 2 decimal places. (2 marks)
(d) A second tower of exactly the same height is built at P. Without calculating its angle of elevation from O, say whether it would be greater or smaller than 6°, and why. (2 marks)
4. Justify
3 marks
Q4.1 A student says: "Radial surveying is a waste of effort. Just walk round the boundary with a tape and measure the four sides." Give two reasons this does not work, one practical and one mathematical.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The error and what the check proves
The error is in the last angle only. ∠DOA crosses north, so the two bearings cannot be subtracted directly; 265° is the reflex angle on the far side of the pair of lines [1].
Correctly, ∠DOA = (360° − 300°) + 035° = 60° + 35° = 95° [1]. The survey then closes: 85 + 85 + 95 + 95 = 360° [1]. Note that 265° + 95° = 360°, which is the signature of this particular mistake.
What the check proves. It confirms that your subtractions are right. It does not confirm that the surveyor read the bearings correctly, because any four bearings, taken in order round the circle and subtracted properly, will always total 360°. So a closing survey rules out your arithmetic as the problem; it says nothing about the compass [1].
Q2.1 (a), Two possible angles
Area = ½ ab sin C, so 2194.35 = ½(64)(80) sin C = 2560 sin C [1], giving sin C = 2194.35 / 2560 = 0.857168.
C = inverse sin(0.857168) = 59°, and because sin θ = sin(180° − θ), the second possibility is 121° [1].
Q2.1 (b), Which one, and how you know
The angles at O must total 360°. The other three give 85 + 92 + 124 = 301° [1], so the missing angle is 360 − 301 = 59°. The correct answer is therefore 59°, and 121° is rejected [1].
This is the one place in a radial survey where the closing check does real mathematical work rather than just catching arithmetic. The area rule alone genuinely cannot distinguish 59° from 121°; the geometry of the full survey can.
Q3.1 (a), Angle POR
Both bearings are measured from north at O, so ∠POR = 168° − 018° = 150° [1].
Q3.1 (b), Distance PR
Two sides and the angle between them, so the cosine rule: PR² = 96² + 110² − 2(96)(110) cos150° = 21316 − 21120(−0.866025) = 21316 + 18290.46 = 39606.46 [1]. So PR = 199.01 m [1].
The angle at O is well obtuse, so the correction term is added and PR comes out close to the sum 96 + 110 = 206 m, which is what a nearly straight line through O would give.
Q3.1 (c), Height of the tower at R
The tower is vertical, so its height sits in a right-angled triangle whose base is the horizontal distance OR = 110 m and whose angle at O is the 6° elevation [1].
height = 110 tan6° = 110 × 0.105104 = 11.56 m [1].
Q3.1 (d), The second tower at P
Greater than 6° [1]. P is only 96 m from O while R is 110 m away, and the two towers are the same height. In tan(elevation) = height ÷ horizontal distance, the numerator is unchanged and the denominator is smaller, so the tangent is larger and therefore so is the angle [1].
Put in words, the same object seen from closer up sits higher in your field of view.
Q4.1, Why walking the boundary is not enough
Practical reason. A boundary is often not walkable. Rivers, dams, scrub, fences, buildings and steep ground all sit on the line you would have to tape, and the whole appeal of a radial survey is that the surveyor never leaves one spot [1].
Mathematical reason. Four side lengths do not determine a quadrilateral. Unlike a triangle, a four-sided figure with fixed sides can flex, and its area changes as it does, so four tape measurements alone cannot give you the area at all [1]. The radial survey fixes the shape by cutting it into triangles, and a triangle is rigid once you have two sides and the angle between them [1].