Page 53 - ACE YR IGCSE A TOP APPR' TO MATHS

PT Volume = Area of PSTW × TU
tan ∠PST = = 205.456 × 10
TS = 2054.56 m 3 [4]
tan ∠PST = 6 (d) WU = TU + TW
2
2
2
25
= 10 + 20
2
∠PST = 13.5° [4] WU = 22.361 2
5 (a) sin ∠PQS = sin ∠SPQ Let the angle of elevation = x
PS SQ 2.25
sin ∠PQS = 4.1 sin 32° tan x = 22.361
6.2
x = 5.75°
∠PQS = 20.51° [3] 7 (a) ∠DAC = 230° – 180° [4]
(b) Bearing of R from Q = 50°
= 180° + (148° – 20.51° – 72°) ∠ADC = 180° – 50° × 2
= 235.49° [2] = 80°
(c) PQ = SQ Bearing of C from D
sin ∠PSQ sin ∠SPQ = 50° + 80°
PQ = 6.2 sin 127.49° = 130° [3]
sin 32° (b) AC = AD + DC – 2(AD)(DC) cos ∠ADC
2
2
2
= 9.28 km [3] 1Rights Reserved.
= 58 + 58 – 2(58)(58) cos 80°
2
2
AC = 74.56 m
(d) QR = SR (c) AC = AB + BC – 2(AB)(BC) cos ∠ABC [4]
2
2
2
sin ∠QSR sin ∠SQR 74.56 = 52 + 75 – 2(52)(75) cos ∠ABC
2
2
2
QR = 7.5 sin (180° – 127.49°) ∠ABC = 69.2°
sin 72°
= 6.257 km Area of ABC = 1 (AB)(BC) sin ∠ABC
Total distance 2
= PQ + QR + RP = 2 (52)(75) sin 69.2°
= 9.28 + 6.257 + 7.5 + 4.1 = 1822.91 m 2 [6]
= 27.137 km [5] (d) (i) Shortest distance
(e) ∠SRQ = 180° – 32° – 72° – 20.51° = AD × sin ∠DAC
= 55.49° = 58 × sin 50°
Shortest distance = 44.43 m [2]
= QR × sin ∠SRQ (ii) Area of the whole garden
= 6.257 × sin 55.49 1
= 5.156 km [4] = 2 (58)(58) sin 80° + 1822.91
(f) Area = 3500 m 2 [3]
1
Penerbitan Pelangi Sdn Bhd. All 7π = ∠POR × 2 × π × 6
=
(PQ)(QR) sin ∠PQR
(a)
8
2 3 360
= 1 (9.28)(6.257) sin (72° + 20.51°) ∠POR = 70° [2]
2 (b) (i) Area of sector OPQR
= 29 km 2 [2] = 70 × π × 6
2
6 (a) Area ΔWST = 1 (TS)(TW) sin ∠STW 360
2 = 7π [2]
= 1 (18)(20) sin 40° (ii) Area of ΔOPR
2 1
= 115.7 m 2 [2] = 2 (6)(6) sin 70°
(b) SW = ST + TW – 2(ST)(TW) cos ∠STW = 16.91 cm 2 [2]
2
2
2
SW = 18 + 20 – 2(18)(20) cos 40° (c) Area of shaded segment PQR
2
2
2
SW = 13.13 m = 7π – 16.91
SW = PS + PW – 2(PS)(PW) cos ∠SPW = 5.081
2
2
2
13.13 = 14 + 17 – 2(14)(17) cos ∠SPW Percentage = 5.077 × 100%
2
2
2
∠SPW = 48.96° [8] π × 6 2
(c) Area of trapezium PSTW = 4.49% [3]
= Area of ΔWPS + Area of ΔWST 1
1 (d) Area of ΔABC = 2 (AB)(AC) sin ∠BAC
= (14)(17) sin 48.96° + 115.7
2 1
= 205.456 m 2 10 = 2 (5)(5) sin ∠BAC
∠BAC = 53.13°
Answers 173

Answers.indd 173 15/03/2022 11:08 AM