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Warm up Notes Preliminary Activity Activity Surveying For Fun Back 1. Find the area of the following triangles: a) Area = ½absinC = ½ x 9 x 8 x sin115o = 32.6cm2 b) Area = ½absinC = ½ x 6.9 x 4.8 x sin108o = 15.7cm2 c) Area = ½absinC = ½ x 19 x 18 x sin105o = 165.2cm2 d) Area = ½absinC = ½ x 6.9 x 7.8 x sin118o = 23.8cm2 Back Offset survey - this was looked at in Year 11 as a field diagram. The longest diagonal of a field is used as a traverse (to cross) line, and the perpendicular distances from this line to each corner of the field are measured - these are called offsets. The plane table radial survey and the compass bearing radial survey methods result in similar diagrams. The cosine rule is used to find the length of each boundary and hence the perimeter. The area of each part is found using A = ½absinC. The main difference in the methods is that the angles must be measured from the plane table diagram using a protractor, while the angles in the compass survey are calculated from the compass bearings. The perimeter of the plane table survey is usually found using measurement and calculation while the compass survey uses the cosine rule to find the perimeter. Back Next A Example 1: Plane table radial survey The following information was the result of a plane table radial survey. i) Find the perimeter of each field ii) Calculate the area of each field B D i) Use the cosine rule to find the length of the missing side. c2 = a2 + b2 - 2ab cosC AB2 = 542 + 482 - 2 x 54 x 48 x cos110o AB2 = 6993.03.... AB = √6993.03.... AB = 83.6m BC = 57.2m; CD = 47.7m; ∴ Perimeter = 257m DA = 68.5m ii) The area of each part is found by using A = ½absinC ΔAPB = ½ x 54 x 48 x sin110o = 1217.8m2 ΔBPC = 943.1m2 ΔCPD = 730.2m2 ΔDPA = 995.6m2 ∴ Area = 3886m2 C E Back A Example 2: The compass radial survey A field was surveyed, all measurements in metres. i) Calculate the perimeter using the cosine rule. ii) Calculate the area. D B C i) <APB = 125o - 53o = 72o <BPC = 195o - 125o = 70o <CPD = 269o - 195o = 74o <DPE = 345o - 269o = 76o <EPA = 35o + 53o = 68o AB2 = 7.92 + 12.12 - 2 x 7.9 x 12.1 x cos 72o AB2 = 149.7.... AB = √149.7... AB = 12.2m BC = 8.6m; CD = 8.9m; DE = 13.4m; 14.1m ∴ Perimeter = 57m ii) A = ½absinC ΔAB = ½ x 7.9 x 12.1 x sin72o Δ AB = 45.5m2 BC = 26.4m2; ΔCD = 26.3m2; 72.9m2 ∴ Area = 220m2 Δ EA = ΔDE = 48.6m2; ΔEA = Back Complete exercise 5-10 Questions 1, 2, 4, 7, 8, 10, 12, 14 41.7% 56.3% 75.7% Back