Page 50 - ACE YR IGCSE A TOP APPR' TO ADD MATH

21. (a) sin ∠OAE = OE
OA y
OA = 10
7
sin 1 36 π 2
= 17.434 cm 3

AD = 17.434 + 10
= 27.434 cm

DC = 27.434 × 1.222
= 33.52 cm
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
[3]
(b) tan ∠OAE = OE
AE x
AE = 10 0 π 2π
7
tan 1 36 π 2 [3]
= 14.281 cm 2. 1 – sec A
sec A
Area of EOBA 1 1
= 2 × Area of triangle OEA = 1 – cos A
= 2 × 3 10 × 14.281 4 cos A 1
2
= 142.81 cm = cos A – cos A
2
2
Area of major sector EOB = cos A – 1
cos A
2
1
= 1 (10) 2 25 π –sin A
2
2 18 =
= 218.166 cm sin A
2
tan A
Area of sector DCA = −sin A tan A
1
2
= 1 (27.434) 2 7 π [4]
2 18 3. 4 sin q + 2 cos q = 0
= 459.75 cm 4 sin q = −2 cos q
2
Area of shaded region sin q = – 1
= 459.75 − 142.81 − 218.166 cos q 2
= 98.78 cm tan q = – 1
2
[4] 2
q = 153.43° or 333.435°
[4]
9 Trigonometry cos 2B 1
2
4. + = 3
sin 2B sin 2B
2
2
1. (a) and (b) cos 2B + 1 = 3 sin 2B
2
2
1 1 −sin 2B + 1 = 3 sin 2B
2
2
xy =
2 sin 2B = 1
2
y = 1 2
2x 1
8 solutions sin 2B = ± √2
[3] −360° < 2B < 360°
2B = ±45°, ±135°, ±225°, ±315°
B = ±22.5°, ±67.5°, ±112.5°, ±157.5°
[4]

Cambridge IGCSE
TM
190 Ace Your Additional Mathematics

Answers Add Math.indd 190 14/03/2022 12:29 PM