educational communications and technology
DESCRIPTION
EDD 5161E. Educational Communications and Technology. Instructor: Dr. Lee Fong Lok Mr. Tam Tat Sang. Group : 9 Student: Lau Pui Shan 98029680 Cheung Chung Yee 98038180. 三角學的應用. 三角學. A. 平面 (1). (a) 三不共線的點. 例 : 棋盤上的棋子. A. B. C. A. 平面 (2). (b) 一直線和線外一點. - PowerPoint PPT PresentationTRANSCRIPT
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Instructor: Dr. Lee Fong Lok Mr. Tam Tat SangGroup : 9
Student: Lau Pui Shan 98029680
Cheung Chung Yee 98038180
![Page 3: Educational Communications and Technology](https://reader036.vdocuments.us/reader036/viewer/2022062309/56814f0e550346895dbca1ff/html5/thumbnails/3.jpg)
三角學的應用平面兩平面的夾角一線與一面線與面的交角斜坡上的最大斜率
三角學
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(a) 三不共線的點
B
例 : 棋盤上的棋子
A
C
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(b) 一直線和線外一點 例 : 船的桅桿和帆
底部的外端
A
L
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(c) 兩相外直線
L1
例 : 支撐風箏的兩條竹枝
L2
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(d) 兩平行直線
L2
例 : 火車路軌
L1
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(a) 互相平行
1
2
例 : 房間相對的兩幅牆
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(b) 相交與一線
1
例 : 計算機的封套
2
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(c) 重疊
1
例 : 兩張相疊的紙張
2
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1. 圖中 VABCD 為一以方形為底的角錐體 , 試指出平面 VCD 和ABCD 的交角。
C
A
B
D
V
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2. 圖中 ABCDEFGH 為一長方體, P 為 AD上的一點,指出平面PFG 和 EFGH 的交角。
A B
EF
G
DC
H
P
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(a) 若一直線一不相容的平面相交 , 則交 點為一點。
L
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(b) 若一直線與一平面互不相交 , 則它們 互相平行。
L
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(c) 若一直線垂直於一平面上任何一直線 , 而那些直線均穿過該直線的垂足 , 則該直線垂直於該平面。
L
L2
L3
L1
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3. 右圖中 , L 垂直於平面 , 而 O 為 L 與的交點 , 試標出所有以 O 為頂的直角
L
C
B
AO
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4. 右圖中 ,
ABCDHEFG 為一正方體 , 試標出所有以A 為頂的直角。
H E
G F
C B
D A
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一點在一平面上的投影是由該點至平面所作垂線的垂足 , 因此 , A’ 為 A 點在平面上的投影。
A
A’
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線段 A’B’ 為線段 AB 在上的投影。
A
B
A’ B’
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將圖中線段 AB 平行移下 , 直至 A 點到達平面 , 那麼線段 AB 與投影 AB’ 之間的角度便是 AB 與平面的交角。
A B
A’
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5. 右圖中由三個長方形平面 ABCD , ABFE , DEFC 組成 , 試標出線 BE 與平面 ABCD的交角。
E F
D
A B
C
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6. 圖中 ABCDEFGH 為一長方體 , 試標出線EB 與平面 ABCD 的交角
E H
F
A
D
G
B
C
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已知兩相交平面 , 一傾斜、一為水平面。在平面上任何垂直於交線的線便稱為最大斜率線。
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圖中 , XYAB 和 XYCD 為兩相交平面 , XYCD 傾斜而 XYAB 為水平面 , 則 PQ為最大斜率線。
要留意的是 > 。此外 , 最大斜率與水平面的交角同時也
是兩平面之間的交角。
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D C
X Y
AB
P
Q
R
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圖中所示的長方形盒中 , AB=BC=4 ,AQ=3 。P S
QR
AB
3
4
4C
D
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(a) 求 BD 的值。
解 :
BD= 22ADAB (畢氏定理)
= 2424
=32
P S
QR
AB
3
4
4C
D
在
ABD 中 ,
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(b) 求線 BP 和平面 ABCD 的交角的值。
解 :
tan =DBPD
=323=27.90 (1d.p.)
線BP和平面ABCD的交角是27.90
P S
QR
AB
3
4
4C
D
在 BDP 中 ,
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一山坡的最大斜率與水平線的交角為 150 , 小徑長 100m, 與最大斜率線成 60o 。
E F
C
BA
D100m
150600
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(a) 求 BF 的值。
解 :
cos600 =BEBF
= 100BF
BF = 100 cos600
= 50m
C
BA
D100m
150600
E F在 BEF 中 ,
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(b) 求 FC 的值。
解 :
sin 150= BFFC
= 50FC
= 12.9m (1 d.p.)
C
BA
D100m
150
E F
C
BA
D 150600
在 BCF 中 ,
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(c) 求小徑頂與水平面的垂直高度。
解 : 小徑頂至水平面的垂直高度
= DE
= FC
= 12.9m (1 d.p.)C
BA
D100m
150600
E F
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(d) 求小徑與水平面的斜角的值。
解 :
sin = BEDE
= 1009.12
= 7.40 (1d.p.)小徑與水面的斜角的值是7.40
C
BA
D100m
150600
E F在 BDE 中 ,