O-LevelMathematicsTrigonometry4024/21 · Oct/Nov 2025 · Q209 Marks
The diagram shows a triangular field, ABC. AB = 65 m, AC = 75 m and angle CAB = 125°. [Figure of triangle ABC]
- (a)Show that BC = 124.3 m, correct to 1 decimal place.[3]
- (b)Calculate angle ABC.[3]
- (c)Calculate the shortest distance from A to BC.[3]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- 20(a) √65² +75²-2×65×75cos125 M2 M1 for 65² + 75² - 2×65×75×cos125 A1 124.26 to 124.27
- 20(b) 29.6 or 29.59 to 29.64 [3] M2 for 75 sin125 / 124.3 oe or [cos... = ] 65² +124.3²-75² / 2×65×124.3 oe or M1 for 75/sinB = 124.3/sin125 oe or 75² = 65² + 124.3² - 2 × 65 × 124.3 × cosB oe
- 20(c) 32.1 or 32.09 to 32.15 [3] M2 for distance/65 = sin(their ABC) oe or 1/2 × 75 × 65 sin125 = 1/2 × 124.3 × h or M1 for recognition that shortest distance is perpendicular to BC
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