Page 52 - ACE YR IGCSE A TOP APPR' TO MATHS

14 (a) (i) Cross-section area = 331.765 + 932.803
1 360 = 1264.6 cm 2 [4]
= 5 × × 5 × 5 × sin
2 5 (d) Bearing of S from P = 90°+ ∠SPQ
= 59.441 cm 2 [3] = 90° + 112.558°
(ii) Volume = 59.441 × 20 = 202.6° [1]
= 1188.82 cm 3 [1] 2 (a) sin ∠QRP = sin ∠QPR
(b) Mass =1188.82 × 13.5 g QP QR
= 16049.07 g sin ∠QRP = 100 sin 55
= 16.05 kg [2] 110
(c) 1 m × 0.9 = 0.9 m ∠QRP = 48.132°
3
3
= 900000 cm 3 ∠PQR = 180° – 55° – 48.132°
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Number of prisms = 76.868° [4]
900000 (b) (i) Bearing of R from Q
=
1188.82 = 180° – (360° – 345°)
= 757.05 = 165° [1]
∼ 757 prisms [3] (ii) Bearing of Q from P
15 Volume of the box = 180° – (360° – 165° – 76.868°)
= (1.4 cm × 2 × 4) × (1.4 cm × 2 × 4) × (1.4 cm × 2) = 61.9° [2]
= 351.232 cm 3 (c) PR = QR
Volume of 16 balls sin PQR sin QPR
4 110 sin 76.868
= 16 × × π × 1.4 3 PR =
3 sin 55
21952 = 130.77 m
= π cm 3
375 Distance
Percentage of the balls that occupied the box = 130.77 × cos (345° – 270° – 48.132°)
21952 = 116.65 m [5]
= π ÷351.232
375 (d) Area = 1 (PQ)(QR) sin ∠PQR
1 2
= π% [6]
6 = 1 (100)(110) sin 76.868
2
6 6 Trigonometry = 5356.17 m 2 [2]
3 (a) ∠PQR = 360° – 90° – 240°
Trigonometry = 30° (shown) [1]
(b) (i) PR = PQ + QR – 2(PQ)(QR) cos ∠PQR
2
2
2
2
2
2
1 (a) SR = SQ + QR – 2(SQ)(QR) cos ∠SQR PR = 11.5 + 20 – 2(11.5)(20) cos 30°
2
2
2
75 = 45 + 45 – 2(45)(45) cos ∠SQR PR = 11.57 km [4]
2
2
2
∠SQR = 112.885° [4] (ii) sin ∠PRQ = sin 30°
sin ∠SPQ sin ∠PQS 11.5 11.57
(b) =
SQ PS ∠PRQ = 29.8°
sin ∠SPQ = ( 45 sin 38° ) Bearing of P from R
= 180° – 90° – 30° – 29.8°
30
∠SPQ = 180° – 67.442° = 30.2° [5]
= 112.558° (c) Total time used
PQ PS = 5.5 h – 20 min – 20 min
= 29
sin ∠PSQ sin ∠PQS = h
30 sin 29.442 6
PQ = Total distance travelled
sin 38°
= 23.95 cm [7] = 11.5 km + 20 km + 11.57 km
= 43.07 km
(c) Area of PQRS Speed of running
= Area of ΔPQS + Area of ΔSQR
1 1 = 43.07 km
= (PQ)(QS) sin ∠PQS + (SQ)(QR) 29
2 2 h
sin ∠SQR 6
1 1 = 8.91 km/h [3]
= (23.95)(45) sin 38° + (45)(45) 4 TS = TR + RS 2
2
2
2 2 2 2
sin 112.885° = 15 + 20
TS = 25 cm
Cambridge IGCSE
TM
172 Ace Your Mathematics

Answers.indd 172 15/03/2022 11:08 AM