Upload
qhayyum-hakiem
View
240
Download
0
Embed Size (px)
DESCRIPTION
math
Citation preview
1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1.
FE
H
J
G
M
LK
FGJK is an inclined plane.
Answer: C
2.
P Q
RS
T
O
The orthogonal projection of line TQ onto plane PQRS is OQ.
Answer: B
3.
P
M
QR
T
The angle between line TM and the plane PQR is ∠TMR.
Answer: A
4.
E F
GH
J K
LM
The angle between line JG and the base is ∠JGE.
Answer: C
5.
E
FG
H
JK
LM
The angle between line LE and plane EHMJ is ∠MEL.
Answer: D
6.
P Q
RS
N
T
The angle between line TQ and the base PQRS is ∠TQN.
Answer: A
CHAPTER
20 Lines and Planes in 3-Dimensions
CHAPTER
2
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
7.
E
F
G
H
M
The angle between plane GEH and plane GEF is ∠HMF.
Answer: A
8.
E F
GH
K
J M
L
The angle between plane FGJK and plane EFGH is ∠EFK or ∠HGJ.
Answer: C
9.
P Q
RS
T U
VW
The angle between plane PQV and the base PQRS is ∠RQV.
Answer: D
Paper 2
1.
E F
G
9 cm12 cm
8 cm
H x
J
tan x = 8–––12
x = 33.7°
The angle between line JG and plane EFGH is 33.7°.
2.
A B
C10 cm
9 cm D
E
x
tan x = 9–––10
x = 41°59′
The angle between line EC and plane ABCD is 41°59′.
3.
E F12 cm5 cm
10 cm
GHx
K
J M
L
HF 2 = HE 2 + EF 2
HF = 52 + 122
= 13 cm
tan x = JH––––HF
= 10–––13
x = 37°34′
The angle between line JF and the base EFGH is 37°34′.
4.
S R
Qx
7 cm7 cm
24 cm
P
T
QS2 = SR2 + RQ2
QS = 72 + 72
= 98 cm
3
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
tan x = TQ––––QS
= 24––––98
x = 67°35′
The angle between line ST and the base PQRS is 67°35′.
5.
12 cm
E F
H
7 cm
G
sin ∠HFG = 712
∠HFG = 35°419
The angle between line HF and the base EFG is 35°419.
6.
A Bx
C
8 cm
7 cm6 cm
D
E H
FG
tan x = 6—8
x = 36°52′
The angle between plane BCEF and the base ABCD is 36°52′.
7.
7 cm3 cm
4 cmA
BCx
F
E
D
tan x = 4—3
x = 53°8′
The angle between plane ABFE and the base ABCD is 53°8′.
8.
5 cm
10 cm6 cm
P
Q
Mx
R
T
PM 2 = PQ2 – QM2
PM = 102 – 62
= 8 cm
tan x = TP––––PM
= 5—8
x = 32°
The angle between plane TQR and the base PQR is 32°.
Paper 1
1. J M
H
GN
F
EK
The angle between plane FHJ and plane KHJ is ∠KMF.
Answer: D
4
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
2.
E F
GH
K
J M
L
The angle between plane KGJ and the base EFGH is ∠JGH.
Answer: B
3.
R
L
M
N
P
TQ
The angle between line TN and the base LMNP is ∠TNL.
Answer: A
4.
M N
P
TS
R U
Q
The angle between the plane SNP and the plane PNTU is ∠SNT.
Answer: D
5.
P
Q
R
S
T
The angle between line TS and the base PQRS is ∠TSQ.
Answer: B
6.
E F
V
GHU
K
T
J M
L
The angle between the plane FTL and the plane FGML is ∠TUV.
Answer: D
7.
S T
QRM
L P
N
The angle between plane PRSN and plane LMNP is ∠SNM.
Answer: A
Paper 2
1.
A B10 cm
3 cm
3 cm
7 cm CD
E H
F
M
N L
G
(a) The angle between the plane MBC and the base ABCD is ∠MLN.
(b) tan ∠MLN = MN––––NL
= 7––10
∠MLN = 35°
5
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
2.
W 12 cm
10 cm
5 cm
15 cm
X
YZ
P
M
S R
Q
(a) The angle between line XM and plane WZSP is ∠XMW.
(b) MW 2 = PM2 + PW2
MW = 52 + 152
= 250 cm
tan ∠XMW = WX––––MW
= 12–––––250
∠XMW = 37.2°
3. E H
C
BA
F DG
10 cm
15 cm
6 cm
7 cm
(a) The angle between the line CE and the plane ADEF is ∠CED.
(b) tan ∠CED = CDED
= 107
∠CED = 55°
4. (a)
A
FD
B
CG
HE
10 cm
7 cm
4 cm
The angle between the plane DFG and the plane ABGF is ∠AFD.
(b) tan ∠AFD = 104
∠AFD = 68°12′
5. F
A x
E
CB 16 cm
8 cm
7 cm
D
tan x = ED––––AD
= 7–––16
x = 23°38′
The angle between plane EAB and the base ABCD is 23°38′.
6. (a)
F
E
C
BA
D
7 cm
8 cm
6 cm
The angle between the plane BCEF and the base ABCD is ∠ABF or ∠DCE.
(b) tan ∠ABF = 68
∠ABF = 36°52′
7. J K
Gx
FE 15 cm8 cm
9 cm
H
EG2 = EF 2 + FG2
EG = 152 + 82
= 17 cm
tan x = KG––––EG
= 9–––17
x = 27°54′
The angle between line EK and the base EFGH is ∠KEG and ∠KEG = 27°54′.
6
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
Paper 1
1.
E
FG
H
KJ
The angle between line FK and plane GHK is ∠GKF.
Answer: D
2.
E F
GH
NJ K
LM
The angle between line FN and plane EHMJ is ∠ENF.
Answer: A
3.
E
F
M
G
H
The angle between line HG and plane EFH is ∠GHM.
Answer: C
4.
P Q
RS
T
The angle between line QT and plane PST is ∠PTQ.
Answer: D
5.
P Q
RS
T U
VW X
The angle between plane PWV and plane RSWV is ∠PWS.
Answer: A
6.
PX Q
R
V
The angle between the base PQR and plane VPQ is ∠RXV.
Answer: B
7.
P Q
RS
T U
The angle between plane QST and plane RSTU is ∠RSQ.
Answer: B
8.
E F
GH
JM
V
The angle between plane VFG and the base EFGH is ∠VJM.
Answer: C
7
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
9.
E F
GHK
YJ
X
M L
The angle between plane XFGM and plane JKLM is ∠GML.
Answer: D
10. T U
S
YP Q
X
R
The angle between plane UPS and the base PQRS is ∠UYX.
Answer: A
11.
E F
GHP
R Q
LM
KJ
The angle between plane PGL and plane HGLM is ∠RQP.
Answer: B
12.
J K
PQ
LM
N
The angle between plane NKL and plane NJM is ∠PNQ.
Answer: C
Paper 2
1.
A 6 cm
12 cm
3 cm
4 cmB
CD
M
V
x
BM 2 = BC2 + CM 2
BM = 42 + 32
= 5 cm
tan x = VM––––BM
= 12–––5
x = 67°23′
The angle between line VB and the base ABCD is ∠VBM and ∠VBM = 67°23′.
2.
DC
V
AB
M
12 cm
12 cm
13 cm
x
BD2 = BC2 + CD2
BD = 122 + 122
= 288 cm
BM = 288–––––2
= 8.485 cm
tan x = VM––––BM
= 13–––––8.485
x = 56°52′
The angle between line VB and the base ABCD is ∠VBM and ∠VBM = 56°52′.
8
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
3.
A B
CD
V
12 cm
9 cm
8 cm
x
BD2 = BC2 + CD2
BD = 92 + 122
= 15 cm
tan x = VD––––BD
= 8–––15
x = 28°4′
The angle between line VB and the base ABCD is ∠VBD and ∠VBD = 28°4′.
4.
A
D
C B
V
10 cm
5 cm
5 cm
9 cm
x
BD2 = AB2 – AD2
BD = 102 – 52
= 75 cm
tan x = VB––––DB
= 9––––75
x = 46°6′
The angle between line VD and the base ABC is ∠VDB and ∠VDB = 46°6′.
5.
E F12 cm
5 cm6 cm GH
K
J M
L
x
FH 2 = HE2 + EF2
FH = 52 + 122
= 13 cm
tan x = JH––––FH
= 6–––13
x = 24°479
The angle between line JF and the base EFGH is 24°479.
6.
A
B10 cm
7 cm
8 cm
6 cm 7 cm
4 cm
CJ
D
EH
GF
x
DJ = 6 cm AJ2 = AD2 + DJ 2
AJ = 82 + 62
= 10 cm
tan x = EJ––––AJ
= 7–––10
x = 35°
The angle between line EA and the base ABCD is 35°.
9
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
7.
L MT
9 cm 6 cm
3 cm
5 cm
N
RQ
P x
PT2 = PN2 + NT 2
PT = 92 + 32
= 90 cm
RT 2 = RN2 – TN2
RT = 52 – 32
= 4 cm
tan x = RT––––PT
= 4––––90
x = 22°52′
The angle between line RP and the base LMNP is 22°52′.
8.
A B
C
24 cm
7 cm
D
E H
F G
x
tan x = EH––––HC
= 24–––7
x = 73°44′
The angle between line HC and plane BCEF is ∠ECH and ∠ECH = 73°44′.
9.
4 cm
6 cm TMU
S
P Q
RNx
tan x = PS––––NS
= 4––6
x = 33°41′
The angle between plane PMN and plane RSUT is ∠PNS and ∠PNS = 33°41′.
10.
BA
8 cm6 cm
7 cm
D C
V
x
tan x = VC––––DC
= 7––8
x = 41°11′
The angle between plane VAD and the base ABCD is ∠VDC and ∠VDC = 41°11′.
11.
BMA
8 cm
9 cm
6 cm
CN
D
E F
GH
x
tan x = GN––––MN
= 9—8
x = 48°22′
The angle between plane ABG and the base ABCD is 48°22′.
10
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
12.
8 cm6 cm
15 cm
E FN
GH
J M
KL
x
HF 2 = HE2 + EF2
HF = 62 + 82
= 10 cm
HN = 10–––2
= 5 cm
tan x = JH––––HN
= 15–––5
x = 71°34′
The angle between plane JEG and the base EFGH is 71°34′.
13.
A E
M
B
D
9 cm
9 cm15 cm
x
AM 2 = AB2 – BM2
AM = 152 – 92
= 12 cm
tan x = BM––––AM
= 9–––12
x = 36°52′
∠BAE = 2x = 73°44′
The angle between plane ABD and plane AED is ∠BAE and ∠BAE = 73°44′.
14.
A B
CD
MON
12 cm
5 cm
V
x
tan x = OM––––VO
= 5–––12
x = 22°37′
∠MVN = 2x = 45°14′
The angle between plane VAD and plane VBC is ∠MVN and ∠MVN = 45°14′.
15.
E F13 cm
13 cm
13 cm
G
O
H
J
K L
M
Nx
HF 2 = HE 2 + EF 2
HF = 132 + 132
= 338 cm
OH = 338–––––2
= 9.192 cm
NH = 13–––2
= 6.5 cm
tan x = NH––––OH
= 6.5––––––9.192
x = 35°16′
The angle between plane NEG and the base EFGH is 35°16′.
11
Mathematics SPM Chapter 20
© Penerbitan Pelangi Sdn. Bhd.
16.
L M
T
U
O
PN
QR
12 cm
4 cm
5 cm
x
QT 2 = QN 2 – TN 2
QT = 52 – 42
= 3 cm UO = QT = 3 cm
OT = 12–––2
= 6 cm
tan x = UO––––OT
= 3—6
x = 26°34′
The angle between plane UMN and the base LMNP is 26°34′.