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1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1.
8 mP Q
T
θ
12 m
tan θ = 12–––8
θ = 56.3°The angle of elevation of T from P is 56.3°.
Answer: A
2.
12 m
h m
R
QP41.5°
tan 41.5° = h–––12
h = 12 × tan 41.5° = 10.62 m
Answer: B
3.
9.5 m
H
GF
38.4°
38.4°
tan 38.4° = FH––––9.5
FH = 9.5 × tan 38.4° = 7.53 m
Answer: B
4.
9 m
6 m
x
x
R
Q P
tan x = 6—9
x = 33°41′
The angle of depression of P from R is 33°41′.
Answer: D
5.
20 m26.5°
E
G
F
tan 26.5° = FG––––20
FG = 20 × tan 26.5° = 9.97 m
Answer: B
6.
62 m
48°U
T
V
tan 48° = 62––––UV
UV = 62–––––––tan 48°
= 55.8 m
Answer: B
CHAPTER
19 Angles of Elevation and Depression
CHAPTER
2
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. P
RQ 12 m
14 m
tan ∠QRP = 1412
∠QRP = 49°249
The angle of elevation of P from R is 49°249.
Answer: C
2. G
EF
7 m
N
65°
65° x
tan 65° = 7–––FN
FN = 7–––––––tan 65°
= 3.264 m FE = 2 × 3.264 = 6.528 m
tan x = 7––––––6.528
x = 47°
The angle of elevation of G from E is 47°.
Answer: D
3.
U S
TW
V
The angle of elevation of V from T is ∠WTV.
Answer: D
4.
30 m
50 m
P
XY
tan ∠PYX = 3050
∠PYX = 30°589
The angle of depression of point Y from P is 30°589.
Answer: A
5.
32 m
x m42°
tan 42° = 32x
x = 32tan 42°
= 35.54
Answer: C
6.
5°10° 3.5 m
PP
T Y X
Let P = Eye level
tan 5° = 3.5TX
TX = 3.5tan 5°
= 40.01 m
tan 10° = 3.5TY
TY = 3.5tan 10°
= 19.85 m
XY = TX – TY = 40.01 – 19.85 = 20.16 m
Answer: A
3
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
7.
12 m
24 m
38°
x
X
TV
U
S
x
tan 38° = UV––––12
UV = 12 × tan 38° = 9.375 m SX = 24 − 9.375 = 14.625 m
tan x = SX––––UX
= 14.625––––––12
x = 50°38′The angle of depression of U from S is 50°38′.
Answer: B
8.
F H
E Gθ
Answer: A
9.
30 m28°
h
Q
P
tan 28° = h–––30
h = 30 × tan 28° = 15.95 m
Answer: B
10.
35°
35°
5 m
Nur
Angle of depression
Cat
Answer: B
11.
8 m
26°Q
P
R
tan 26° = 8––––RQ
RQ = 8––––––tan 26°
= 16.4 m
Answer: D
Paper 1
1.
20 m
62°YX
W
sin 62° = YW––––20
YW = 20 × sin 62° = 17.66 m
Answer: C
4
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
2. V
X
U6 m
30° 42°
Y
tan 42° = UV––––6
UV = 6 × tan 42° = 5.4024 m
tan 30° = UV––––XU
XU = UV–––––––tan 30°
= 5.4024–––––––tan 30°
= 9.36 m
Answer: C
3.
4 mP Q
T
8 m26°
R
tan 26° = QT
––––8
QT = 8 × tan 26° = 3.9019 m
tan ∠TPQ = QT
––––4
= 3.9019–––––––4
∠TPQ = 44°17′
The angle of elevation of T from P is 44°17′.
Answer: B
4. H
EF G13 m
Angle of depression
9 m48°
tan 48° = HF––––9
HF = 9 × tan 48° = 9.9955 m
tan ∠HGF = HF––––13
= 9.9955–––––––13
∠HGF = 37°33′
The angle of depression of G from H is 37°33′.
Answer: C
5. M
P Q
Nx
Angle of depression
9 m
5 m
4 m 4 m
8.5 m
tan x = 5––––8.5
x = 30.47°
The angle of depression of N from M is 30.47°.
Answer: A
6. T
S
R
P Q
The angle of elevation of T from P is ∠TPQ.
Answer: D
7.
R
T
3 m
18.5°5.5 mP Q
tan 18.5° = RP––––5.5
RP = 5.5 × tan 18.5° = 1.84 m PT = 3 + 1.84 = 4.84 = 4.8 m
Answer: D
5
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
8.
28°
R
U
T
2 m
16 mV
tan 28° = RU––––16
RU = 16 × tan 28° = 8.51 m TR = 8.51 − 2 = 6.51 m
Answer: B
9.
L N
P
Q
The angle of elevation of the window cleaner, P, from the man, L, is ∠NLP.
Answer: A
10. H
Fy
xG
E
6 m9 m
3 m
15 m
tan x = 9–––15
x = 30°58′
tan y = 3–––15
y = 11°19′x − y = 30°58′ − 11°19′ = 19°39′The difference between the angles of elevation of H and G from E is 19°39′.
Answer: B
11.
16°10°P
T
R
Q24 m
tan 10° = RQ
––––24
RQ = 24 × tan 10° = 4.232 m
tan 16° = TQ
––––24
TQ = 24 × tan 16° = 6.882 m TR = TQ − RQ = 6.882 − 4.232 = 2.65 mThe distance between the two birds is 2.65 m.
Answer: C
12.
22°
22°
Q N
TP
M
30 m 30 m
120 m
tan 22° = MT––––PT
MT = PT × tan 22° = 120 × tan 22° = 48.48 m MN = 48.48 + 30 = 78.48 = 78.5 m
Answer: D
13.
R
T
V
U
Q
20.8 m
51.4 m
Angle of depression
350 m
x
6
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
VT = 51.4 − 20.8 = 30.6 m
tan x = VT––––UT
= 30.6––––350
x = 5°The angle of depression of U from V is 5°.
Answer: B
14. R
T
P Q
S1.8 m1.8 m
40°
6 m
tan 40° = RT–––ST
RT = ST × tan 40° = 6 × tan 40° = 5.03 m RP = 5.03 + 1.8 = 6.83 mThe height of the bird from the ground is 6.83 m.
Answer: D
15. H
E
GF
6.5 m51.6° 72°
72°
tan 51.6° = HF––––6.5
HF = 6.5 × tan 51.6° = 8.2 m
tan 72° = HF––––FG
FG = HF–––––––tan 72°
= 8.2–––––––tan 72°
= 2.66 m
Answer: C
16.
Y X
V
Z
10 m
26 m12.5°
XZ 2 = XY 2 − YZ 2
XZ = 262 − 102
= 24 m
tan 12.5° = ZV––––XZ
ZV = XZ × tan 12.5° = 24 × tan 12.5° = 5.32 m
Answer: C
17.
P Q
H
63°760 m
tan 63° = HQ
––––760
HQ = 760 × tan 63° = 1492 mThe height of the helicopter from the ground is 1492 m.
Answer: C
18.
G
H
E
F
x
32.5°
120 m
80 m
tan 32.5° = HG––––80
HG = 80 × tan 32.5° = 50.966 m
7
Mathematics SPM Chapter 19
© Penerbitan Pelangi Sdn. Bhd.
tan x = HG––––FG
= 50.966–––––––120
x = 23.01°
The angle of depression of F from H is 23.01°.
Answer: A
19. T
R Q P35°
80 m
55°
35°55°
tan 35° = TR––––80
TR = 80 × tan 35° = 56.02 m
tan 55° = TR––––RQ
RQ = TR–––––––tan 55°
= 56.02–––––––tan 55°
= 39.23 m
PQ = PR − RQ = 80 − 39.23 = 40.77 = 40.8 m
The distance between the two cars is 40.8 m.
Answer: C
20.
P Q R
HJ
The angle of depression of Q from H is ∠JHQ.
Answer: A
21.
Q
T
RP
16 m
34 m28.5°
QR2 = PR2 − PQ2
QR = 342 − 162
= 30 m
tan 28.5° = TQ
––––16
TQ = 16 × tan 28.5° = 8.687 m
tan ∠TRQ = TQ
––––QR
= 8.687–––––30
∠TRQ = 16°9′
The angle of elevation of T from R is 16°9′.
Answer: B
22.
P
R Q
T
25 m
4.5 m
7 m
V
PQ2 = QR2 − RP2
PQ = ABBBBBB252 − 72
= 24 m
PT = 24–––2
= 12 m
tan ∠VPT = VT––––PT
= 4.5–––12
∠VPT = 20°33′
The angle of elevation of V from P is 20°33′.
Answer: C