Page 17 - 1202 Bank Soalan Matematik Tambahan Tingkatan 5

BAB 1 5. A (b) OL = 18 kos 41.44°
Tinggi dari PQ
Kertas 1 B = 18 – 18 kos 41.44°
15 cm
Bahagian A C θ = 4.51 cm
1
1. A 9 cm 9. 60 saat = 33— putaran
3
O 100 1 5
3
OC
9
3
60
j jθ (a) —— = — 1 saat = —— × —– = — putaran
10
5
θ OB 5 3 (a) — × 2π = —–π radian per saat
O B OC = — × 15 9 9
j 5
P = 2j + jθ = 23.4 = 9 cm (b) j = 20 cm
10
2
2
2
2
(a) 2j = 23.4 – 5.4 AB = 15 + 15 – 2(15) kos θ Laju = —–π × 20
15 + 15 – 10
2
2
2
j = 9 cm kos θ = ——————– 9 –1
(b) 9θ = 5.4 2(15) 2 = 69.81 cm s –1
θ = 0.6 rad θ = 38° 57ʹ = 0.698 m s
∴ ∠AOB = 0.6 rad θ = 0.68 rad 10.
2
2
2
(b) AC = 15 + 9 – 2(15)(9) kos 38° 57ʹ
2.
AC = 9.8 cm D C
Perimeter rantau berlorek 5 50 cm
K = 6 + 15(0.68) + 9.8 A —π rad B
9
= 26 cm 24 cm
θ
11 cm O
L 6.
O (a) Luas yang dilalui oleh pengelap
1
(a) —(11) θ = 160 cermin
2
2
2 5
1
1
2 5
 2
 2
θ = 2.64 rad O = —(74) —π – —(24) —π
∴ ∠KOL = 2.64 rad 2 9 2 9
(b) 12 cm = 4 276.06 cm 2
C (b) Perimeter kawasan yang dilalui oleh
pengelap cermin
A B
5
5
 2
 2
(a) Luas ΔAOB = 72 = 24 —π + 2(50) + 74 —π
9
9
1
—(12)AB = 72 = 271.04 cm
2
j
AB = 12 cm 11. (a) Perimeter rantau berlorek
π
2
 2
∠AOB = 45° = — rad = 10 —π + 10 sin 72°
Lilitan, KL = 11(2.64) 4 5
= 29.04 cm (b) Luas sektor minor AOC + (10 – 10 kos 72°)
1
2πj = 29.04
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π

 4 2
= 28.99 cm

= —(12) —
2
j = 4.62 cm 2 (b) Luas rantau berlorek
= 18π cm 2 1
2 2
 2
3. = —(10) —π
B 2 5
7. P 1
2
A 12 – —(10 kos 72°)(10 sin 72°)
j cm R 11 1 = 48.14 cm 2
10 2
12.
θ rad T
9 3
O O
AB
θ
(a) sin — = —— 8 4 Q π
2 2j 7 6 5 U –– rad
6
θ
2j sin — = AB j
2 1 O
(b) ∠AOB = 1.2 rad = 68.75° ∠POQ = — × 360° = 120° 2
3
2
Beza ∠ROP = — × 30° + 30° = 50° (a) Luas = 20 cm
1
1
2 π
2  2
= Panjang lengkok AB – Panjang 3 20 = —j — – —j sin 30°
2
6
2
perentas AB Jumlah sudut = 170° = 2.97 rad 2 π sin 30°
2 2

= 5(1.2) – 2(5) sin 34.38° 8. 20 = j —– – ———–
12
= 0.353 cm 20
2
4. P j = ——————–
sin 30°
π
2
O —– – ———–
12
41.44° j = 41.17 cm
18 cm
A (b) Panjang perentas UT
15 cm L R = 2(41.17) sin 15°
O
— π rad B = 21.31 cm
3 30 cm (a) Panjang lengkok = 100 cm,
Q Perimeter tembereng
π
 2
(a) Perimeter j = 18 cm = 21.31 + 41.17 —
6
2
2
 3 2
 3 2
= 15 —π + 2(30) + 45 —π jθ = 100 = 42.87 cm
100
θ = ——
18
= 185.66 cm θ = 5.56 rad
(b) Luas kertas
1 2 2 1 2 2 θ = 318.56°
 3 2
 3 2
= —(45) —π – —(15) —π
2 2
= 600π cm 2
99
10A_1202 BS MT Tg5.indd 99 10/01/2022 4:42 PM