Page 54 - ACE YR IGCSE A TOP APPR' TO MATHS

Area of shaded region (b) y = cos x
53.13 y
= × π × 5 – 10
2
360
= 1.59 cm 2 [4] 1
PS
9 sin ∠PRS =
RS
9
∠PRS = sin –1 ( ) 0 90° 180° 270° 360° x
18
= 30° –1
PS
cos ∠PSQ =
QS [2]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
9 (c) y = tan x
QS =
cos 30° y
= 10.392
∠QSR = 180° – 90° – 30° – 30°
= 30°
Area of ΔQSR x
1 0 90° 180° 270° 360°
= (SQ)(SR) sin ∠QSR
2
1
= (10.392)(18) sin 30°
2 [2]
= 46.76 cm 2 [7] 2 (a) 360° – (210° – 180°)
sin ∠QPR sin ∠PRQ = 330° [2]
10 (a) =
QR PQ (b) sin x = 2
3
∠QPR = sin –1 ( 53.7 sin 47° ) x = 41.81°, 180° – 41.81°
40.2
= 77.68° x = 41.81°, 138.19° [3]
∠PQR = 180° – 47° – 77.68° 3 (a) cos x = 0.5
= 55.32° [4] x = 60°, 360° – 60° [3]
= 60°, 300°
(b) Area = 1 (PQ)(QR)sin ∠PQR (b) cos x = –0.5
2 x = 180° – 60°, 180° + 60°
= 1 (40.2)(53.7) sin 55.32° x = 120°, 240° [3]
2
= 887.61 cm (c) cos x = –0.25
2
= 0.0888 m 2 [2] x = 180° – 75.522°, 180° + 75.522°
AB x = 104.48°, 255.52° [3]
11 sin ∠ACB = (d)
AC y
AB = 10 sin ( 180 – 90 ) 2
2
= 7.071 1

A = Area of sector AEC – (Area of sector ABC – 0 90° 180° 270° 360° x
Area of ΔABC) –1
) (
45
= ( 360 × π × 10 – 1 × π × 7.071 – 7.071 2 ) –2 [2]
2
2
4
2
= 25 cm 2 [7] 4 (a) x = 360° – 50°
= 310° [2]
(b) –0.75 ≤ y < – 1
Trigonometry graph y
1 (a) y = sin x 1
y

1
0 90° 180° 270° 360° x

x
0 90° 180° 270° 360° –1

–1 [2]
[2]
Cambridge IGCSE
TM
174 Ace Your Mathematics

Answers.indd 174 15/03/2022 11:08 AM