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(b) ln y = ln e –1.5x + 1.35
2
y = e −1.5x 2 · e 8 Circular Measure
1.35
y = 3.86r
–1.5
[3] 1. (a) ∠QOR = π – 2 1 2
π
15 – 5 6
19. (a) m = = 5 2π
3 – 1 =
5 = 5(1) + c 3 [1]
c = 0
2
2
lg y = 5x (b) RSQ = OR
lg y = lg 10 sin ∠QOR sin ∠RQO
2
5x
2
2
y = 10 RSQ = 5
2
5x
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5x 2 2π π
y = 10 sin 3 sin 6
2
5
a = , b = 0 RSQ = 5√3 cm
2
[5] Perimeter
(b) When x = 1 = 5 × 2π + 5√3 × π
2 3 2
5( 2 ) = 24.08 cm
1 2
y = 10 2
5 Area
= 10 1 5√3 2 1 2 2π 1 2π
8
= 4.22 = 2 1 2 2 (π) – 3 2 (5) 1 2 – (5)(5) sin 3 4
3
2
[2] 75 25 25√3
(c) lg 3 = 5x = 8 π – 1 3 π – 4 2
2
2
x = 2 lg 3 = 25 π + 25√3
2
5 24 4
2
x = ± 2 lg 3 = 14.1 cm
5 [6]
= ±0.44
[2] 2. (a) OQ = 6 cm ÷ 2
= 3 cm
20. (a) m = 7 – (–3) = –5 QR = OR – OQ
2
2
2
0 – 2
2
2
7 = –5(0) + c = 6 – 3
c = 7 QR = 3√3 cm
1
e = –5 1 2 + 7 PT = PO + OT
y
x
= PT – PO
OT
e
ln = ln (–5x + 7) = 8 cm − 6 cm
–1
y
1
y = ln 7 – 5 x 2 = 2 cm
QR
[5] tan q = QO + OT
5
(b) 7 – = 0
x = 3√3
7 = 5 3 + 2
x q = 0.805 rad
x . 5
7 [2] [3]
(b) RT = RQ + QT 2
2
2
1 15 2
2
2
(c) y = ln 7 – 5 = (3√3) + 5
6 RT = 2√13 cm
= ln 5 SR = ST – RT
[2] = 8 – 2√13
5
4
(d) e = 7 – x cos ∠ROQ = 3
5 = 7 – e 6
4
x ∠ROQ = π
1
x = 5 3
7 – e 4 [2]
Answers 185
Answers Add Math.indd 185 14/03/2022 12:29 PM