O-LevelMathematicsTrigonometry4024/22 · May/June 2025 · Q229 Marks
The diagram shows a triangular field ABC. AB = 150m, BC = 187m and angle ABC = 112°. [Figure showing triangular field ABC]
- (a)Fencing is needed for the perimeter of the field. Fencing is sold in rolls of length 20 m. Calculate the number of rolls of fencing needed.[5]
- (b)Calculate the shortest distance from B to AC.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [5] B4 for 30.9 or 30.85 to 30.86 OR M2 for [AC=]√(150²+187²-2×150×187×cos112) oe or M1 for [AC² =] 150²+187²-2×150×187×cos112 oe After M1 earned allow A1 for 78484[....] M1dep for 617 to 617.2 or their AC + 150 + 187 M1dep for their perimeter ÷ 20
- [4] M3 for 150×187×sin(112) / their AC oe or M2 for 1/2 ×150×187×sin112 = 1/2 ×their AC×h or for [sin C =] (150 sin112) / their AC oe or for [sin A =] (187 sin112) / their AC oe or M1 for 1/2 ×150×187×sin112 oe or for indicating AC and shortest distance are perpendicular
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