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Mathematics Form 3 PT3 Model Paper
Total distance (ii) ∠FBC = 180° – 150°
(b) Average speed =
Time = 30°
[1.5 × (10 × 60) + 5 000 + x] 7 000
= ∴ x + 130° + 30° = 180°
10 + 32.5 + 17.5 60 x = 20°
x = 1 100m
∠DEB + x = 180°
∴ Ken walked for 1 100 m. ∠DEB = 180° – 20°
= 160°
10.24
(c) (i) tan ∠KJL = y = 360 – 160°
8 = 200°

∠KJL = tan –1 10.24 
8
∠KJL = 52°

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BOOKLET ANS MATH F3.indd 32 03/01/2020 10:21 AM