Simple Course on 3-D

8/18/2019 Simple Course on 3-D

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拔教 Simple ‘A’ Class Tel: 2104 3598

1

Simple Course on 3-D

Lines and Planes:

Plane is parallel to line:

Intersection point of plane & line


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Projection of a line


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Angle between line and plane

The angle between a line and a plane is the angle between the line and its

projection on the plane.

http://www.youtube.com/watch?v=4y9diopip8Q

http://www.youtube.com/watch?v=LjT_-hzsp5M

http://www.youtube.com/watch?v=4y9diopip8Qhttp://www.youtube.com/watch?v=LjT_-hzsp5Mhttp://www.youtube.com/watch?v=LjT_-hzsp5Mhttp://www.youtube.com/watch?v=4y9diopip8Q


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Checkpoint:1.

In the following figure, ABCDEF is a right triangular prism.

Which of the following is the angle between the line EB and the plane

CDEF?

A. DBE

B. CBE

C. BEC

D. BFE

Solution: C


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2. The figure shows a triangular prism.

ABCD and ABFE are rectangles, CF BF and DE AE.

(a)

Find the projection of CA on the plane ABFE.

(b) Find the projection CA on the plane CDEF.

(c) Name the angle between CA and the plane ABFE.(d)

Name the angle between CA and the plane CDEF.

(12 marks)

Solution:(a)

AF

(b)

CE

(c)

CAF

(d)

ACE


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3. The figure shows a right prism with a right-angled triangle as thecross-section. Find the angle between the line BF and the plane

ABCD correct to the nearest degree.

A.

20 B.

30

C.

35

D.

40

E.

50

Solution: A


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4. In Figure 12, PQ is a vertical tower h m high, A and B are two points100 m apart with A, B and Q on the same horizontal ground. AQB = 75° and the angles of elevation of P from A and B are 40° and 60°

respectively.

(a) Express the lengths of AQ and BQ in terms of h. (2 marks)

(b) Find h and QAB. (5 marks)(c) Find the angle between the plane PAB and the horizontal. (4 marks)


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Solution:

(a) tan40°=

AQ = ℎ

40°

=1.19h tan60°= ℎ BQ = ℎ60° = ℎ√ 3 =0.577h

1.19h= ℎ40° ℎ√ 3 =0.577h

(b) 100 = 1.19 + 0.577ℎ − 21.19ℎ0.577ℎ75° 10000=1.39 36ℎ 8 4 . 7 = h

84.7√ 3sin = 10075° a=28.2° (c) By Herone’s formula ∆AQB=2382

2382= 12 100 47.6=H

sin40°=.

60°=.

PA = 131.8 PB = 97.8

By Herone’s formula ∆AQB = 4859 4859 =

100

PM = 97.2

= .. Q=60.6°


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5. A vertical rectangular wall on the horizontal ground, 2 m high and 6m long, runs east and west as shown in the figure. If the sun bears S

45 E at an elevation of 60, find the area of the shadow of the wall

on the ground.

A. 2 3 m2

B.

4 3 m2

C.

4 2 m2

D. 6 m2

E.

2 6 m2

Solution: E


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6. The figure shows a cuboid ABCDEFGH with XH = a, GH = b and

FG = c. If the angle between XH and the plane EFGH is , then cos

=

A.

a b .

B.

a2

b.

C. c2

b.

D.

a

c b 22 .

E.

a2

c b 22 .

Solution: E


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7. In the figure, ABCDEFGH is a cuboid. tan =

A.

3

1.

B. 3

1 .

C.

1.

D. 3 .

E.

3.

Solution: E

http://www.youtube.com/watch?v=r3DZxl4iiV8

http://www.youtube.com/watch?v=r3DZxl4iiV8http://www.youtube.com/watch?v=r3DZxl4iiV8http://www.youtube.com/watch?v=r3DZxl4iiV8


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8. In the figure, ABCDA’B’C’D’ is a rectangular box of base 9 cm by

6 cm and height 6 cm and its lid D’C’QP opens by an angle of 30.

Find the angles between

(a)

BD’ and D’C’CD

(b) BD’ and BCC’B’

(c)

B’P and A’B’C’D’

Give your answers correct to the nearest 0.1.


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Solution:

DB2 = 62 + 92 (Pyth. theorem)

DB2 = 117

(a). Let the angle between be 1

tan 1 =PB

6

1 = 29.0

(b).

Let the angle between be

tan =72

9

= 46.7

(c).

Let the angle between be

PE = 6 sin 30 = 3

tan =04.93

= 18.4

EB’ = 22 9)30cos66( =

9.04


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Angles between 2 planes

If from a point on the line of intersection of two planes, one in each plane,

are at it so found between these lines is defined as the between the

two planes.


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Checkpoint:1.

ABCD is a regular tetrahedron. E, F and G are the mid points of AB,

AC and AD. Write down the angle between plane ABC and ACD

Solution: angle BFD


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2. The base PQRS is a square. A is the mid-point of PQ, VP = VQ =VR = VS. The projection of the point V on the plane PQRS is M,

which is also the intersection of PR and SQ.

(a) Name the angle between the planes VPQ and PQRS.(b)

Name the angle between the base PQRS and the edge

(i)

VP,

(ii)

VS,

(iii) VR

(12 marks)


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Solution:

(a) VAM; VRM

(b)

(i) VPM;

(ii)

VSM(iii)

VRM


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3. In the figure, VABCD is a right pyramid with a square base. Eachedge of the pyramid is of equal length. M and N are the mid-points

of AB and BC respectively. If is the angle between the line VM

and the line VN, then cos =

A.

6

1

B. 2

1

C.

3

2

D.

6

5


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Solution:

C

Join MN.

∵ MVN is the angle between the line VM and the line VN.

∴ = MVN

Let a be the length of the edge of the pyramid.

∵ VA = VB and AM = BM

∴ VM AB (prop. of isos. )

In VMB,22 BMVBVM (Pyth. theorem)

a2

3

2

aa

2

2

Similarly, VN = a2

3.

In BNM,22 BMBNMN (Pyth. theorem)

a2

2

2

a

2

a 22

In VMN, by the cosine formula,

cos =)VN)(VM(2

MNVNVM 222


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3

2

4

32

2

1

4

3

4

3

a2

3a

2

32

a2

2a

2

3a

2

3222


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4.

The figure shows a right-angled triangular prism ABFCDE with BF =

8cm, CD = 30cm and ∠BCF=25°.(a)

Find the angle between the planes AFD and ABCD. (3 marks)

(b) Let X be a point moving from F to C. Describe the change angle

between the planes AXD and ABCD. (1 marks)


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Remarks: FZ is vertical pole.

tan = ZX/YZ; YZ is constant but ZX is max at F so that decrease form

ans in a to 0

https://youtu.be/uZMUX2g_nAo

https://youtu.be/uZMUX2g_nAohttps://youtu.be/uZMUX2g_nAohttps://youtu.be/uZMUX2g_nAo


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Perpendicular line to a plane:

The line PN is normal to the plane .

It can be said to be perpendicular to every line on the plane.


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The figure shows a ○ ABCDEFGH. Which of the following is/are true?

I.

EG is the projection of DG on plane EFGH.

II.

AG is the projection of DG on plane AFGB.

III. DG is the projection of DH on plane DEGB.

A.

I only

B.

II only

C.

I and II only

D. I, II and III

Solution: D


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3. In the figure, PC is a vertical pole standing on the horizontal plane

ABC. If ABC = 90, BAC = 30, AC = 6 and PC = 5, find tan.

A.

5

3

B. 6

5

C. 3

5

D.

35

3

E.

59

3

Solution: C

tan =3

5

30sin6

5

BC

5


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4. If the area of three adjacent faces of a rectangular block are

√ 2, √ 3 and √ 6, then the length of diagonal isA. 2√ 3 B.

3√ 2

C. √ 3

D. √ 6

Ans: D


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Line of greatest slope

A line which is on the inclined plane and is perpendicular to the common

edge of the planes is called line of greatest slope.


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Checkpoint: 1.

Fill in the blanks

(a) CE is perpendicular to the plane ___,

(b) is the angle between the line _ and the plane ___,

(c)

is the angle between the planes ___ and ___,(d)

if ABCD is a horizontal plane, then _ is a line of greatest

slope.


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Solution:(a)

ABCD

(b)

AE, ABCD

(c)

ABEF, ABCD

(d)

BE


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2. Figure 4 shows a geometric model ABCD in the form of tetrahedron.

It is found that ACB = 60, AC = AD = 20 cm, BC = BD = 12 cm

and CD = 14 cm.

(a)

Find the length of AB.

(2 marks)

(b) Find the angle between the plane ABC and the plane ABD.

(4 marks)

(c)

Let P be a movable point on the slant edge AB. Describe howCPD varies as P moves from A to B. Explain your answer.

(2 marks)


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(a) By cosine formula,

AB2 = AC2 + BC2 – 2(AC)(BC)cosACB

AB2 = 202 + 122 – 2(20)(12)cos60

AB = 194 cm

(b) By sine formula,

AB

ACBsin

BC

BACsin

194

60sin

12

BACsin

BAC 36.58677555

Let Q be the foot of the perpendicualr from C to AB.sinBAC =

AC

CQ

CQ 20sin36.58677555

CQ 11.92079121cm

Since ABC ABD, the reqruied angle is CQD.

CQ

CD2

1

2

CQDsin

2CQDsin

0.587209345

CQD 71.91844786

CQD 71.9

Thus, the angle between the plane ABC and the plane ABD is

71.9.

By sine formula,


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194

60sin

20

ABCsinAB

ACBsin

AC

ABCsin

ABC 83.41322445 Let Q be the foot of the perpendicular from C to AB.

sinABC =BC

CQ

CQ 12sin83.41322445

CQ 11.92079121cm

Since ABC ABD, the required angle is CQD.

587209345.02

CQDsin

CQ

CD2

1

2

CQDsin

CQD 71.91844786

CQD 71.9


71.9.

By sine formula,

19460sin

12BACsin

AB

ACBsin

BC

BACsin

BAC 36.58677555

Let Q be the foot of the perpendicular from C to AB.

sinBAC =AC

CQ

CQ 20sin36.58677555

CQ 11.92079121cm

By symmetry, we have DQ = CQ.

DQ 11.92079121cmSince ABC ABD, the requrid angle is CQD.

CD2 = CQ2 + DQ2 – 2(CQ)(DQ)cosCQD

142 11.920791212 + 11.920791212

- 2(11.92079121)(11.92079121)cosCQD

CQD 71.91844786

CQD 71.9


71.9.


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The area of ABC

sin)BC)(AC(2

1 ACB

60sin)12)(20(

2

1

360 cm2


360)CQ)(AB(2

1

360)CQ)(194(2

1•

CQ 11.92079121cm

Since ABC ABD, the reqruied angle is CQD.

CQ

CD2

1

2

CQDsin

2

CQDsin

0.587209345

CQD 71.91844786

CQD 71.9


71.9.

(c)


Note thatCP

CD2

1

2

CPDsin

.

Since CP CQ, we have CPD CQD.

Thus, CPD increases as P moves from A to Q and decreases

as P moves from Q to B.

https://youtu.be/ABG7sd17fek

https://youtu.be/ABG7sd17fekhttps://youtu.be/ABG7sd17fekhttps://youtu.be/ABG7sd17fek


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Question about bearing:

East, South, West and North direction will be given in this type of

question.

There are two bearing systems

True bearing

Compass bearing

Two diagrams have to be drawn

Situation Diagram

Radar Diagram


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Checkpoint:

1.

In the figure, points A, B and C are on the same horizontal plane,

CD is a vertical tower of height 50 m. B is due east of the tower

while A is S30W from the tower. The angles of elevation of the

lower from A and B are 30 and 60 respectively. P is a point on AB

and CP AB. Find

(a) the distance between A and B. (5 marks)

(b)

the distance between C and AB, i.e. CP. (4 marks)

(c)

the angle of elevation of D from P. (2 marks)

Give the answers correct to the nearest integer.


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Solution:

(a)

AC =

30tan

50

BC =60tan

50

cos 180 =)BC)(AC(2

ABBCAC 222

AB2 = AC2 + BC2 – 2(AC)(BC) cos 120

AB = 104 m (corl. to nearest integer)

(b) CAPsin

BC

120sin

AB

CAP = 13.89788625

sinCAP =AC

CP

CP = 21 cm (corl. to nearest integer)

(c)

tanCPD =CP

50

CPD = 67 (corl. to nearest integer)


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2. In the figure, OT is a vertical pole of height h m. A boy located at A

is 10 3 m due North of O. He finds that the angle of elevation of T

from him is 25. A girl located at B is 10 m due East of O.

(a).

Find h and hence, find the angle of elevation of T from B.(3 marks)

(b).

The boy walks due East and the girl walks in a bearing N E

with the same speed as the boy. If they meet at point C, find

(i).

OAB and hence, the value of .

(ii).

AB, OC and the angle of elevation of T from C.

(6 marks)

(c).

If now the girl walks back along CB to another point D (notshown in the figure) such that the angle of elevation of T from

D is the same as that from C, find BD.

(3 marks)


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Solution:

(a).

H = 8.08 m; Angle of elevation of T from B = 38.9

(b).

(i). OAB = 30 , = 30

(ii).

AB = 20 m, OC = 10 7 m

Angle of elevation of T from C = 17

(c).

As the angle of elevation is the same

OD & OC must be same & OBD is 60 as he is walking

along BC

(10 7 )2 = 102 + BD2 -2(10)BD cos 120

600 = BD2 + 10BD

BD2 + 10BD – 600 = 0

BD = 30 or – 20 (rejected)


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Figure 8

In Figure 8, OT is a vertical tower of height h metres and O, P and Q are points on the same horizontal plane. When a man is at P, he finds that the

tower is due north and that the angle of elevation of the top T of the tower

is 30. When he walks a distance of 500 metres in the direction N50E to

Q, he finds that the bearing of the tower is N70W.

(a)

Find OQ and OP. (3 marks)

(b)

Find h. (2 marks)

(c) Find the angle of elevation of T from Q, giving your answer

correct to the nearest degree. (2 marks)

(d)

(i)

If he walks a further distance of 400 metres from Q in a

direction NE to a point R (not shown in Figure 8) on

the same horizontal plane, he finds that the angle of

elevation of T is 20. Find OQR and hence write

down the value of to the nearest integer.

(ii)

If he starts from Q again and walks the same distance of

400 metres in another direction to a point S on the same

horizontal plane, he finds that the angle of elevation of T

is again 20. Find the bearing of S from Q, giving youranswer correct to the nearest degree. (5 marks)

94 IB 14


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Solution:

(a)

POQ = 70,

PQO = 180 – 70 – 50

= 60

考慮△POQ

By sine rule,

50sin

OQ

60sin

OP

70sin

500

OQ =

70sin

50sin500

= 408m (407.6m)

OP =

70sin

60sin500

= 461m (460.8m)

(3 marks)正弦公式

Csin

c

Bsin

b

Asin

a

(b)

h = 460.8 × tan30

= 266 (266.04)(2 marks)

考慮△OPT


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(c)

In△OQT,

tanTQO =6.407

04.266

TQO = 33.1

= 33 (correct to the nearest degree)

(2 marks)

(d)

(i)

Denote and in the figure.

In△ORT.

OR = 20tan

04.266

= 730.95

By cosine rule, in△OQR,

cosOQR =)400)(60.407(2

)95.730(400)60.407( 222

OQR = 129.66

= 130

So, = 130 – 70= 60

先求OR ，再利用餘弦公式求OQR 及，然後再利用△OQR 和△OQS全

等的資料，求出，從而得出由Q測 S的方位角。

(ii)

For△OQR △OQS,

OQR = 130

= 130 – 20 – 90 = 20

So, the bearing of S from Q is S20E

(or 160)

(5 marks)

(1) 注意△OPQ是一個直角三角形。

(2) 同學在回答 (d) (ii)時要考慮△OQR 和△OQS全等的關係。


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Shadow question

Light shines on vertical pole at an angle

The ray is depend on the light sources

The area of shadow can varies

Generally, the object length ≠ shadow length


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Checkpoint:1.

A vertical rectangular wall on the horizontal ground, 2 m high and 6

m long, runs east and west as shown in the figure. If the sun bears

S45E at an elevation of 60, find the area of the shadow of the wallon the ground.

A.

2m34

B. 2m24

C.

2m6

D.

2m62

Solution: D


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2. triangular plate ABC is erected vertically on the horizontal plane andis right-angled at A. AB = 5 m. AC = 10 m. The sunlight shines along

SQ as shown in the figure. The angle of elevation of B from Q is 30.

Given QAC = 60.

(a). Find AQ and QC. (4 marks)

(b).

Find the area of the shadow of the plate. (2 marks)

(c). If now the plate is rotated anti-clockwise through an angle of

90, find the area of the shadow of the plate. (2 marks)

(d). Find the position of the plate (within the range of the included

angle between AC and AQ) such that the shadow of the pate isa maximum. Hence, find the maximum area of the shadow.

(4 marks)


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Solution:

(a).

AQ = 5 3 m, QC = 9.40 m

(b).

37.50 m2

(c).

21.65 m2

(d). 75100 = 5 m

Area = 43.3 m2


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3. In the figure, ABC is a triangular metal sheet in which AB = AC = 2

m and BC = 2 m . It is placed with its edge AB on the horizontal.

The edges AC and BC make angles and with the horizontal

respectively. The metal sheet casts a shadow ABD under the sun. CD

is a vertical line in the figure.

(a).

Show that cos CAB =4

3.

Hence find the area of the metal sheet.

(b). Show that sin = 2 sin .

(c).

Let = 30

(i).

Find the length of AD.(ii). Using the result in (b), find CBD and the length of BD.

Hence determine ADB and the area of the shadow ABD.


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Solution:

(a). In CAB , by cosine law22 = 22 + 22 – 2(2)(2) cos CAB

2 = 8 – 8 cos CAB

cos CAB =4

3

Area =2

1 (2)(2) sin CAB =

2

7 m2

(b). In ACD, sin =2

CD

CD = 2 sin

In CBD, sin =2

CD

CD = 2 sin

Hence, 2 sin = 2 sin

sin = 2 sin (c).

(i). cos =2

AD ; cos30 =

2

AD

23 =

2AD , AD = 3

(ii).

sin = 2 sin

= 2 sin 30 =2

2

= 45

BD = 2 sin 45 = 1

In ADB, by cos law,

22

=2

3 + 12

= 2( 3 )(1) cos ADBcos ADB = 0

ADB = 90

Notes: You may use converse of Pythagoras Theorem

Area =2

1(1) 3 =

2

3 m2


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4. Figure 9 shows a triangular road sign ABC attached to a vertical poleOAB standing on the horizontal ground. The plane ABC is vertical

with OA = 2m, AB = 0.6 m, AC = 0.7 m and BC = 0.8 m. D is a point

on the horizontal ground vertically below C and is due north of the

foot O of the pole.

The sun is due west. When its angle of elevation is 30, the shadow

of the road sign on the horizontal ground is ABC.

(a)

Find the lengths of OA and AB. (3

marks)

(b) Calculate BAC and hence find the length of OD. (4 marks)

(c) Find the area of the shadow ABC.

(2 marks)

(d)

If the angle of elevation of the sun is less than 30,

(i) state whether the shadow of AB is longer than, shorter

than, or equal to AB in (a); and hence

(ii) state with reasons whether the area of the shadow of the

road sign ABC is larger than, smaller than, or equal to

that of ABC in (c).

(3 marks)


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Solution:

(a)

OA = m3230tan

2

AB = OB – OA

30tan

2

30tan

6.2

3236.2

= 0.6 m3 (3 marks)

考慮△AOA

(b) In△ABC,

cosBAC =

)7.0)(6.0(2

8.07.06.0 222 =

4

1

BAC = 75.5

OD = 0.7 × sin75.5

= 0.678m (4 marks)餘弦公式

cosA = bc2

ac b 222


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(c) Area of shadow ABC

=2

1× (AB) × (OD)

= )36.0(2

1× 0.678

= 0.352m2 (2 marks)

(d)

For the angle of elevation is less then 30,

(i) the shadow of the triangle will be increased.

(ii)

Since the length OD remains unchanged while the length

of AB will increase, so the area of the shadow of the

road sign ABC will be larger. (3

marks)

若仰角少於 30， (i) 陰影長度會增加。

(ii)

OD長度保持不變故陰影面積會增加。

(1) 在回答 (c) 部時要注意OD是三角形的高。

(2)

在 (d) 部時注意OD不受太陽的仰角所影響。


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5. Figure 12(a) shows a triangular parking sign ABC. It is made oftransparent plastic plate with a symbol “P”. The plate ABC is

attached to a vertical pole OAB standing on the horizontal ground,

where vertex C is due north of O. D is a point on the horizontal

ground vertically below C.When the sun is due east and its angle of elevation is 40, the

shadow of the parking sign ABC on the horizontal ground is

A1B1C1.

It is given that OA = 3 m, AB = 0.6 m, BC = 0.7 m and CA = 0.5 m.

(a)

Find the area of ABC. (3

marks)

(b)

Find the length of OD.

(2 marks)

(c)

Find the length of A1B1. Hence, find the area of A1B1C1.

(4 marks)

(d)

Now the sun is due west and its angle of elevation is 40. Suppose

A2B2C2 is the shadow of the parking sign on the horizontal

ground.

i. In Figure 12(b), sketch the image A2B2C2.

ii. State with reason whether the area of A2B2C2 is

larger than, smaller than or equal to the area of

A1B1C1.

(2 marks)


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Solution:

(a) s =2

5.07.06.0 = 0.9 1

By Heron’s formula

Area of ABC= cs bsass cm2

= 5.09.07.09.06.09.09.0 m2 1M

= 0.146969 m2

= 0.147 m2 (cor. to 3 sig. fig.) 1A

(3)

(b)

2

1(AB)(OD) = Area of ABC

OD =

6.0

146969.02 m

1M

= 0.489898 m

= 0.490 m (cor. to 3 sig. fig.) 1A

(2)

(c) In OBB1,

OB1 =40tan

OB

=

40tan

6.03m =

40tan

6.3m

In OAA1,

OA1 =

40tan

3

40tan

OA m

A1B1 = OB1 – OA1 1

= 0.715052 m

= 0.715 m (cor. to 3 sig. fig.) 1A

Area of A1B1C1

=2

1(A1B1)(OD)

1M

=2

1(0.715052)(0.489898) m2

= 0.175151 m2

= 0.175 m2 (cor. to 3 sig. fig.) 1A


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(4)

(d)

(i)

1A

(ii)

Area of A2B2C2 is equal to the area of A1B1C1

because A2B2C2 is the image of reflection of A1B1C1

with the axis of reflection OD.

1

(2)


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55

Frustum Question

Frustum must be cut parallel to the base.

We must use similar figure to solve this type of question

V = (

)

= V ()

Volume of frustum = V1 - V2


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56

Checkpoint:1.

Figure 5(a) shows a right pyramid VABCD with a square base, where

VAB = 72. The length of a side of the base is 20 cm. Let P and Q

be the points lying on VA and VD respectively such that PQ is

parallel to BC and PBA = 60. A geometric model is made bycutting off the pyramid VPBCQ from VABCD as shown in Figure

5(b).

(a)

Find the length of AP. (2 marks)

(b)

Let be the angle between the plane PBCQ and the base ABCD.

i.

Find .

ii.

Let be the angle between PB and the base ABCD. Which

one of and is greater? Explain your answer.

(5 marks)


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57

Solution:1.

(a) By sine formula, we have

APBsin

AB

PBAsin

AP

)7260180sin(

20

60sin

AP

AP = 23.30704256cm

AP 23.3cm

Thus, the length of AP is 23.3cm.

(b)

(i)

Let S be the foot of the perpendicular from P to AD.

PS

= APsinPAD

23.30704256sin72

22.1663147cm

AS

= APcosPAD

23.30704256cos72

7.202272239cm

By sine formula, we have

APBsin

AB

PABsin

PB

)7260180sin(

20

72sin

PB

PB 25.59545552cm

Let T be the foot of the perpendicular from P to BC.

PT2 = PB2 – AS2

PT2 (25.59545552)2 – (7.202272239)2

PT 24.56124219cm

Note that = PTS.

By cosine formula, we have

cos =)ST)(PT(2PSSTPT

222

)20)(56124219.24(2

)1663147.22()20()56124219.24(cos

222

= 58.59703733

58.6

(ii)

Let X be the projection of P on the base ABCD.Then, we have = PBX.

1M

1A

1M

1M

1M

1A

1M

r.t. 23.3 cm

-----------------|

either one|

------------------

r.t. 58.6


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58

Note that PB > PT.

sin

PB

PX

PT

PX

= sinPBX

= sin

Since and are acute angles, is greater than .1A

------

--(6)

f.t.


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59

2.

Figure 7 shows a rectangular plane ABCD which is inclined at 30

to the horizontal plane ABEF, where E and F are vertically below C

and D respectively. Suppose DBF = , AFB = 60 and BD = 500m.

(a)

Express BF and DF in terms of .

(2 marks)

(b)

(i)

By considering AFD, express AF in terms of . (Give

your answer in surd form if necessary.)

(ii)

By considering AFB, express AF in terms of .

(iii)

Hence, or otherwise, find the value of correct to 3

significant figures.

(6 marks)

(c)

Find the angle of elevation, as observed from B, of the top of a

vertical pole erected at D and of height 50 m. (3

marks)


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60

Solution:a)

BF = 500co

DF = 500sin

b)

(i)

AF

DF = tan30

3

1

AF

sin500

AF = 3500 sin

(ii) cos60 =cos500

AF

AF = 500cos60cos (iii) cos60cos500sin3500

cos2

1sin3

tan =32

1

1.16 c)

BF = 500cos16.1 DF = 138.675

= 480.384

Total H = 138.675 + 50 = 188.675

tan =384.480

675.188

= 21.4

http://www.youtube.com/watch?v=B6jdf5sqB3k

http://www.youtube.com/watch?v=B6jdf5sqB3khttp://www.youtube.com/watch?v=B6jdf5sqB3khttp://www.youtube.com/watch?v=B6jdf5sqB3k


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61

Paper folding question

Two diagrams have to be drawn

Plane Diagram (Before folding)

3-D Diagram (After folding)

New points must be indicated by ’ , e.g. new A A’, new B B’


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62

Checkpoint:1.

In the figure, ABCD is a square paper. E and F are midpoint of AB

and DC respectively. EF intersects the diagonal BD at O. The paper

b then folded along EF such that AE⊥BE and DF⊥ FC. ∠BOD =

A.

120°

B. 150°

C. 135°

D. 90°

Ans: C


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63

2.

Figure 8a

Figure 8b

In Figure 8a, ABCD is a thin square metal sheet of side three metres.

The metal sheet is folded along BD and the edges AD and CD of thefolded metal sheet are placed on a horizontal plane II with B two

metres vertically above the plane II. E is the foot of the perpendicular

from B to the plane II. (See Figure 8b)

(a)

Find the lengths of BD, ED and AE, leaving your answers in

surd form. (3 marks)

(b)

Find ADE. (3 marks)

(c) Find the angle between BD and the plane II. (2 marks)

(d) Find the angle between the planes ABD and CBD. (4 marks)

92 IB 15


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64

Solution:

(a) BD = 22 33

= 3 2 m

ED = 22 2)23(

= 14 m

AE = 22 23

= 5 m

(3 marks)

考慮△ABD，△BDE及△ABE，分別計算 BD，ED及 AE的值。

(b)

cosADE =

)14)(3(2

5143 222

ADE = 36.7

(3 marks)

cosA = bc2

ac b 22

(c)

The required angle is BDE,

sinBDE =BD

BE

= 23

2

BDE = 28.1

So, the angle between BD and the plane II is 28.1 (2 marks)

考慮△BDE


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65

(d) Let k be the mid point of BD, the required angle is AKC

AK = 22 )2

23(3

=2

23

CK = 22 )2

23(3

=2

23

AC2 = 32 + 32 – 2 (3) (3) cosADC

= 18 – 18 cos 2 × 36.7

= 12.85 76

cosAKC =

2 23 2 3 2( ) ( ) 12.857 62 2

3 2 3 22( )( )

2 2

= – 0.428 6

AKC = 115

So, the angle between the planes ABD and CBD is 115

(4 marks)

(1) 在回答 (a)部時要將答案以根式表示。

(2)

同學要熟習如何找出一個平面和一個直線的交角，兩個平面的

交角。

http://www.youtube.com/watch?v=B6jdf5sqB3k

http://www.youtube.com/watch?v=B6jdf5sqB3khttp://www.youtube.com/watch?v=B6jdf5sqB3khttp://www.youtube.com/watch?v=B6jdf5sqB3k


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2. Figure 4(a) shows a piece of rectangular paper ABCD. It is foldedalong BD and placed on a horizontal table such that the plane BCD is

vertical and the points A, C, D lie on the table as shown in Figure

4(b). Given that AD = a and AB = 2a.

(a)

(i) By the converse of Pythagoras’ Theorem, or otherwise,

prove that ACD is right-angled.

(ii)

Find the angle between the plane ABD and the table.(5 marks)

(b)

(i)

By considering the area of ABD, or otherwise, find the

distance between A and BD in terms of a.

(ii)

In Figure 4(a), let X be the point on BD such that AX

BD and AX produced intersects CD at Y.

(1) By considering the similar triangles AXD and ADY

in Figure 4(a), or otherwise, find AY and DY interms of a.

(2)

Find the angle between the planes ABD and BCD

in Figure 4(b), correct to the nearest degree.

(7 marks)

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