Page 26 - Module & More Matematik Tambahan Tg5

Kertas 2
Tinggi (cm) / Height (cm)
40
3
1. Lakarkan graf y = kos 2x + 1 dan y = kos
2
bagi 0° < x < 360° pada graf yang disediakan
20
di bawah. Kemudian, tandakan ⊗ pada titik
persilangan dua graf tersebut.
Tanah
Masa (s)
x + 1 for
Sketch the graph of y = cos 2x + 1 and y = cos
0
Ground
Time (s)
3
0° < x < 360° on the graph provided. Hence, mark ⊗ on
(a) Nyatakan fungsi trigonometri yang mewakili
the points of intersection of the two graphs.
tinggi pedal basikal itu daripada tanah, H
Jawapan / Answer :
dengan masa, t.

State the trigonometric function that represents the
height of the bicycle pedal from the ground, H and
y
3
y = kos x + 1
time, t.
2
y = kos 3x + 1
2
(b) Berapakah diameter putaran pedal basikal
itu?

What is the diameter of the bicycle pedal rotation?
1
Jawapan / Answer :
x
40 – 20
∴ H = 10 kos 24t + 30

0
(a) a =
= 10
60°
H = 10 cos 24t + 30
2
360°
= 24
b =

15
c = 10 + 20 = 30

2. Rajah berikut menunjukkan tinggi pedal sebuah
(b) Diameter putaran pedal basikal
basikal semasa dikayuh pada laju seragam dalam
Diameter of the bicycle pedal rotation

tempoh seminit. 120° 180° 240° 300° 2 360° x + 1 Matematik Tingkatan 5 Bab 6 Nisbah dan Graf Fungsi Trigonometri
The diagram shows the height of a bicycle pedal while = 40 – 20 = 20 cm
pedalling at a constant speed in a minute.
Matematik Tambahan Tingkatan 5 Bab 1 Sukatan Membulat
Sudut
Sudut KBAT KBAT
Sudut
Sudut
Sudut
Sudut
KBAT
KBAT
KBAT
KBAT
KBAT
Ekstra
6
Populasi satu spesis ikan di suatu tasik diwakili oleh Jawapan / Answer: 1
1. Rajah menunjukkan sebuah bulatan berpusat
(c) Panjang lengkok SP = QR
fungsi P(t) = 4 800 sin t + 6 400, dengan P ialah (a) Populasi maksimum ikan / Maximum fish population

O dan berjejari 10 cm. PQ dan SR adalah dua
Arc length SP = QR
populasi ikan, t ialah masa dalam hari dan 0 < x < 360. = 4 800 + 6 400 = 11 222
5π
perentas yang selari dengan PQ = 10 cm dan
π
 
The population of a species of fish in a lake is represented Populasi minimum ikan / Minimum fish population BAB
=
10
∠SOR = 120°.
3
6
by function P(t) = 4 800 sin t + 6 400, where P is the fish = 4 800 + 6 400 = 1 600 BAB
The diagram shows a circle with centre O and radius
2
2
population, t is the time in days and 0 < x < 360. SR = 10 + 10 − 2(10) kos 120°
2
2
10 cm. PQ and SR are two parallel chords such that
(a) Hitung populasi maksimum dan minimum ikan (b) t = 0, 180, 360
SR = 10 3

PQ = 10 cm and ∠SOR = 120°.
itu. 5π
Calculate the maximum and minimum population of Perimeter = 10 + 2   + 10 3
3
the fish. P (c) 4 800 sin t + 6 400 = 1889
π
Q


(b) Pada hari keberapakah populasi ikan mencapai 6 sin t = –0.9398 = 10 1+ + 3 
R
S
3
–1
400 ekor? a = sin 0.9398
During which days does the fish population reach 6 = 70
O
∴ t = 180 + 70, 360 – 70

400? 2. Rajah menunjukkan satu semibulatan PADB
(c) Hitung nilai-nilai t apabila terdapat 1 889 ekor berpusat P dan satu sektor CAB berpusat C
= 250, 290
Kuiz 6
ikan. dilukis pada satah Cartes.
Calculate the values of t when there was 1 889 fishes. The diagram shows a semicircle PADB with centre P
(a) Cari sudut SOP dalam radian. and a sector CAB with centre C drawn on a Cartesian
Find the angle of SOP in radian. plane.
(b) Tentukan luas rantau berlorek itu.
Determine the area of the shaded region. y
(c) Tunjukkan bahawa perimeter rantau D


berlorek ialah 10 1 + 3 + π  cm. 93 A(0, 6) P © Penerbitan Pelangi Sdn. Bhd.
3
Show that the perimeter of the shaded region is B x

10 1 + 3 + π  cm.
3 C(0, –4)
(a) PQ = 10 + 10 −
2
2
2
2(10) kos POQ Diberi CP = 5 3 unit, cari
2
10 cm Given that CP = 5 3 units, find
10 + 10 − 10 2 P
2
2
kos POQ = Q (a) diameter semibulatan itu.
2(100) S R
120 the diameter of the semicircle.
∠POQ = 60° O 10 cm (b) sudut ACB dalam radian.
120°−60° the angle ACB in radian.
∠SOP =
2 (c) perimeter rantau berlorek
π the perimeter of the shaded region
= 30° = rad
2 (d) luas rantau berlorek.
(b) Luas tembereng PQ /Area of segment PQ the area of the shaded region.
1 π 1
= (10) 2   − (10) sin 60°
2
2 3 2 (a) CA = 10 = CB
2
= 9.06 cm 2 AP = 10 − (25)(3) y
Luas tembereng SR / Area of segment SR = 5 unit
1 2π 1 Diameter AB = 10 unit D
= (10) 2   − (10) sin 120° A(0, 6) E
2
2 3 2 (b) sin ∠PCB = 5 = 1 P
= 61.42 cm 2 10 2 B x
Maka, luas berlorek ∠PCB = 30°
Hence, the shaded area π C(0, –4)
= 61.42 – 9.06 ∠ACB = 60° = rad
3
= 52.36 cm 2
23