Mathematics Standard • Year 12 • Trigonometry • Lesson 8
Radial Surveys in Context
Run a full survey for a real decision: how much fence, how much seed, how much land. Then handle the two things a survey question adds on top of the arithmetic, converting to hectares and pricing the result.
1. The property
A property PQRS is surveyed from a central point O. The surveyor's record reads:
P · 018°T · 96 m Q · 097°T · 74 m R · 168°T · 110 m S · 265°T · 85 m
Q1.1 Find the four angles at O and show that the survey closes. 3 marks
Q1.2 Find the total area of the property, correct to the nearest square metre. 4 marks
Q1.3 Express that area in hectares, correct to 2 decimal places. 1 mark
2. Fencing the boundaries
Same property. 2 marks each
Q2.1 Find the length of boundary QR, correct to 2 decimal places.
Q2.2 Find the length of boundary SP, correct to 2 decimal places.
Q2.3 Boundary QR and boundary SP are built from radial lines of similar length, yet SP is far longer. Explain why, without calculating anything further.
3. Making the decision
3 marks each
Q3.1 Pasture seed is sold in bags that cover 2000 m² each, and bags cannot be split. How many bags are needed to sow the whole property, and how much of the last bag is wasted?
Q3.2 A contractor quotes $38 per metre to fence boundaries QR and SP only. Find the cost, to the nearest dollar. Then explain why the survey had to be done before the quote could be given.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The four angles
∠POQ = 097° − 018° = 79°, ∠QOR = 168° − 097° = 71°, ∠ROS = 265° − 168° = 97° [2].
Wrap-around: ∠SOP = (360° − 265°) + 018° = 95° + 18° = 113°. Check: 79 + 71 + 97 + 113 = 360° [1].
Q1.2, Total area
[POQ] = ½(96)(74) sin79° = 3552 × 0.981627 = 3486.74 m² [1]
[QOR] = ½(74)(110) sin71° = 4070 × 0.945519 = 3848.26 m² [1]
[ROS] = ½(110)(85) sin97° = 4675 × 0.992546 = 4640.15 m² [1]
[SOP] = ½(85)(96) sin113° = 4080 × 0.920505 = 3755.66 m²
Total = 3486.74 + 3848.26 + 4640.15 + 3755.66 = 15730.81, which is 15731 m² [1].
Q1.3, In hectares
One hectare is 10000 m², so 15731 ÷ 10000 = 1.57 hectares.
Q2.1, Boundary QR
QR² = 74² + 110² − 2(74)(110) cos71° = 17576 − 16280(0.325568) = 17576 − 5300.25 = 12275.75 [1], so QR = 110.80 m [1].
Q2.2, Boundary SP
SP² = 85² + 96² − 2(85)(96) cos113° = 16441 − 16320(−0.390731) = 16441 + 6376.73 = 22817.73 [1], so SP = 151.06 m [1].
Q2.3, Why SP is so much longer
The angle at O is what differs, not the radii. ∠QOR = 71° is acute, so cos71° is positive and the correction term is subtracted, pulling QR in below the Pythagoras value. ∠SOP = 113° is obtuse, so cos113° is negative, the correction term is added, and SP is pushed out above it.
Put simply, a wide angle at the central point swings the two corners apart. The radii set the scale of the triangle; the angle at O sets how open it is.
Q3.1, Seed
15731 ÷ 2000 = 7.87 bags [1]. Bags cannot be split, so the farmer needs 8 bags [1]. Those 8 bags cover 16000 m², so the waste is 16000 − 15731 = 269 m² of coverage [1]. Note the rounding goes up here regardless of the decimal, because 7 bags would leave part of the property unsown.
Q3.2, Fencing cost
Total length = 110.7960 + 151.0554 = 261.85 m [1]. Cost = 261.8514 × 38 = 9950.35, which is $9950 to the nearest dollar [1].
Worth noting: adding the two rounded lengths instead gives 261.86 m and a cost of $9951. Rounding before the final step has moved the answer by a dollar, which is Trap 03 from the lesson showing up in money rather than in area.
The survey had to come first because neither boundary was ever measured. The surveyor recorded only bearings and distances from O, so QR and SP existed as calculations before they existed as numbers. Without the cosine rule there is no length to price, and a contractor cannot quote on a fence line nobody has walked [1].