cours

1. 1
‫اﻟﻤﺜﻠﺜﻲ‬ ‫اﻟﺤﺴﺎب‬–‫اﻟﺠﺰء‬2-
‫اﻟﻤﻨﺘﻈﺮة‬ ‫اﻟﻘﺪرات‬
‫ﻋﻠﻰ‬ ‫ﻣﺜﻠﺜﻴﺔ‬ ‫ﻣﺘﺮاﺟﺤﺔ‬ ‫أو‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫وﻗﺮاءة‬ ‫ﺗﻤﺜﻴﻞ‬ ‫ﻣﻦ‬ ‫اﻟﺘﻤﻜﻦ‬
‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬
‫اﻟﺪرس‬‫اﻷول‬‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺪورة‬
‫اﻟﺴﺎﻋﺎت‬ ‫ﻋﺪد‬:15
I-‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬
1-‫اﻟﻤﻌﺎدﻟﺔ‬cosx a=
‫ﻣﺜﺎل‬1‫ﺣﻞ‬
1
cos
2
x x∈ =
‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬
1
:
2
x∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬
‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬
‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬
3
π
‫و‬
3
π
-.
‫أن‬ ‫ﺑﻤﺎ‬2
3
k
π
π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬M
‫و‬2
3
k
π
π+‫ﺑﺤﻴﺚ‬k ∈‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬'M
‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬
1
cos
2
x =‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬2
3
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫إذن‬2 / 2 /
3 3
S k k k k
π π
π π
   
= + ∈ ∪ − + ∈   
   
‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ]
1
2 ;2 cos
2
x xπ π∈ − =
‫ﻋﻠﻰ‬ ‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬
1
cos
2
x =‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬2
3
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;2π π−
‫ﻓﺎن‬2 2 2
3
k
π
π π π− ≤ + ≤‫أو‬2 2 2
3
k
π
π π π− ≤ − + ≤
‫ﻟﺪﻳﻨﺎ‬2 2 2
3
k
π
π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬
7 5
6 6
k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =
‫وﻣﻨﻪ‬
3
x
π
=‫أو‬
5
3
x
π
= −
‫ﻟﺪﻳﻨﺎ‬2 2 2
3
k
π
π π π− ≤ − + ≤‫ﺗﻜﺎﻓﺊ‬
5 7
6 6
k− ≤ ≤‫ﺗﻜﺎﻓﺊ‬1k =‫أو‬0k =
‫وﻣﻨﻪ‬
3
x
π
= −‫أو‬
5
3
x
π
=
‫إذن‬
5 5
; ; ;
3 3 3 3
S
π π π π− − 
=  
 
‫ﺧﻼﺻﺔ‬*‫اﻟﻤﻌﺎدﻟﺔ‬cosx a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺
*cos 1x x∈ =‫وﻓ‬ ‫إذا‬‫آﺎن‬ ‫إذا‬ ‫ﻘﻂ‬/ 2k x kπ∈ =
*cos 1x x∈ = −‫آﺎن‬ ‫إذا‬ ‫وﻓﻘﻂ‬ ‫إذا‬/ 2k x kπ π∈ = +
*‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬] [0;π‫ﺣﻴﺚ‬cos aα =
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬cosx a=‫ﻓﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kα π= − +‫ﺣﻴﺚ‬k ∈
{ } { }2 / 2 /S k k k kα π α π= + ∈ ∪ − + ∈
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2. 2
‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬
( ) ] ]
[ [ 2
3 3
cos cos 2 ;3 cos 2
3 4 2
;2 2cos 3cos 1 0
x x x x x
x x x
π π
π π
π π
   
∈ + = ∈ − − = −   
   
∈ + + =
‫اﻟﺤﻞ‬
*‫ﻧﺤﻞ‬( )cos cos 2
3
x x x
π 
∈ + = 
 
( )cos cos 2
3
x x
π 
+ = 
 
‫ﺗﻜﺎﻓﺊ‬2 2
3
x x k
π
π= + +‫أو‬2 2
3
x x k
π
π= − − +‫ﺣﻴ‬‫ﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬3 2
3
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬
2
9 3
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫إذن‬
2
2 / /
3 9 3
S k k k k
π π
π π
   
= + ∈ ∪ − + ∈   
   
*‫ﻧﺤﻞ‬] ]
3 3
;3 cos 2
4 2
x x
π
π π
 
∈ − − = − 
 
‫أن‬ ‫ﻧﻌﻠﻢ‬
3
cos
6 2
π
=‫وﻣﻨﻪ‬
5 3
cos cos
6 6 2
π π
π
   
− = = −   
   
‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬
3 3
cos 2
4 2
x
π 
− = − 
 
‫ﺗﻜﺎﻓﺊ‬
3 5
2 2
4 6
x k
π π
π− = +‫أو‬
3 5
2 2
4 6
x k
π π
π− = − +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬
19
2 2
12
x k
π
π= +‫أو‬2 2
12
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬
19
24
x k
π
π= +‫أو‬
24
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫ﺣﻴﺚ‬ ‫و‬] ];3x π π∈ −‫ﻓﺎن‬
‫أﺟﻞ‬ ‫ﻣﻦ‬
19
24
x k
π
π= +‫ﻟﺪﻳﻨﺎ‬
19
3
24
k
π
π π π− + ≤≺‫أي‬
19
1 3
24
k− + ≤≺
‫وﻣ‬‫ﻨﻪ‬
43 53
24 24
k− ≤≺
‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬1k = −‫أو‬0k =‫أو‬1k =‫أو‬2K =
‫إذن‬
19 5
24 24
x
π π
π= − = −‫أو‬
19
24
x
π
=‫أو‬
19 43
24 24
x
π π
π= + =‫أو‬
19 67
2
24 24
x
π π
π= + =
‫أﺟﻞ‬ ‫ﻣﻦ‬
24
x k
π
π= − +‫ﻟﺪﻳﻨﺎ‬3
24
k
π
π π π− − + ≤≺‫أي‬
1
1 3
24
k− − + ≤≺
‫وﻣﻨﻪ‬
23 73
24 24
k− ≤≺
‫ﺣﻴﺚ‬ ‫و‬k ∈‫ﻓﺎن‬0k =‫أو‬1k =‫أو‬2K =‫أو‬3k =
‫إذن‬0
24 24
x
π π
π=− + ⋅ =−‫أو‬
23
24 24
x
π π
π= − + =‫أو‬
47
2
24 24
x
π π
π= − + =‫أو‬
71
3
24 24
x
π π
π= − + =
‫إذن‬
5 19 23 43 47 67 71
; ; ; ; ; ; ;
24 24 24 24 24 24 24 24
S
π π π π π π π π 
= − − 
 
*‫ﻧﺤﻞ‬[ [ 2
;2 2cos 3cos 1 0x x xπ π∈ + + =
‫ﻧﻀﻊ‬cos x X=‫ﺗﺼﺒﺢ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬2
2 3 1 0X X+ + =
‫ﻟﻴﻜﻦ‬∆‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻣﻤﻴﺰ‬
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3. 3
2
3 4 2 1 1∆ = − × × =
‫وﻣﻨﻪ‬
3 1 1
4 2
X
− +
= = −‫أو‬
3 1
1
4
X
− −
= = −
‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬
1
cos
2
x = −‫أو‬cos 1x = −
‫ﻟﺪﻳﻨﺎ‬cos 1x = −‫ﺗﻜﺎﻓﺊ‬/ 2k x kπ π∈ = +
‫و‬‫ﺣﻴﺚ‬[ [;2x π π∈‫ﻓﺎن‬2 2kπ π π π≤ + ≺‫أي‬
1
0
2
k≤ ≺‫وﻣﻨﻪ‬0k =‫اذن‬x π=
‫ﻟﺪﻳﻨﺎ‬
1
cos
2
x = −‫أي‬
2
cos cos
3
x
π
=
‫وﻣﻨﻪ‬
2
2
3
x k
π
π= +‫أو‬
2
2
3
x k
π
π= − +‫ﺣﻴﺚ‬k ∈
‫ﺣﻴﺚ‬ ‫و‬[ [;2x π π∈‫ﻓﺎن‬
‫أﺟﻞ‬ ‫ﻣﻦ‬
2
2
3
x k
π
π= − +‫ﻟﺪﻳﻨﺎ‬
2
2 2
3
k
π
π π π≤ − + ≺‫أي‬
5 4
6 3
k≤ ≺‫وﻣﻨﻪ‬1k =
‫إذن‬
2 4
2
3 3
x
π π
π= − + =
‫أﺟﻞ‬ ‫ﻣﻦ‬
2
2
3
x k
π
π= +‫ﻟﺪﻳﻨﺎ‬
2
2 2
3
k
π
π π π≤ + ≺‫أي‬
1 2
6 3
k≤ ≺‫ﻧﺴﺒﻲ‬ ‫ﺻﺤﻴﺢ‬ ‫ﻋﺪد‬ ‫ﻳﻮﺟﺪ‬ ‫ﻻ‬
‫اﻷﺧﻴﺮة‬ ‫اﻟﻤﺘﻔﺎوﺗﺔ‬ ‫ﻳﺤﻘﻖ‬
‫إذن‬
4
;
3
S
π
π
 
=  
 
2-‫اﻟﻤﻌﺎدﻟﺔ‬sin x a=
‫ﻣﺜ‬‫ﺎل‬1‫ﺣﻞ‬
3
sin
2
x x∈ =
‫اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻟﺪﻳﻨﺎ‬
3
:
2
y∆ =‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻳﻘﻄﻊ‬
‫ﻧﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫اﻟﺮﺋﻴﺴﻴﻴﻦ‬ ‫اﻟﻤﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻬﻤﺎ‬
‫هﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬
3
π
‫و‬
2
3 3
π π
π − =.
‫أن‬ ‫ﺑﻤﺎ‬2
3
k
π
π+‫ﺑﺤﻴﺚ‬k ∈‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬
‫ﻟﻠﻨﻘﻄﺔ‬M‫و‬
2
2
3
k
π
π+‫ﺑﺤﻴﺚ‬k ∈‫اﻷﻓﺎﺻﻴﻞ‬ ‫هﻲ‬
‫ﻟﻠﻨﻘﻄﺔ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬'M‫أن‬ ‫ﻧﺴﺘﻨﺘﺞ‬ ‫ﻓﺈﻧﻨﺎ‬
3
sin
2
x =‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬
2
2
3
x k
π
π= +‫ﺣﻴﺚ‬k ∈
‫إذن‬
2
2 / 2 /
3 3
S k k k k
π π
π π
   
= + ∈ ∪ + ∈   
   
‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ ]
3
2 ;3 sin
2
x xπ π∈ − =
‫ﻓﻨﺤﺼﻞ‬ ‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﺨﻄﻮات‬ ‫ﻧﻔﺲ‬ ‫ﻧﺘﺒﻊ‬‫ﻋﻠﻰ‬
3
sin
2
x =‫ﺗﻜﺎﻓﺊ‬2
3
x k
π
π= +‫أو‬
2
2
3
x k
π
π= +‫ﺣﻴﺚ‬k ∈
‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬ ‫أﻧﻨﺎ‬ ‫وﺣﻴﺚ‬[ ]2 ;3π π−
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4. 4
‫ﻓﺎن‬2 2 3
3
k
π
π π π− ≤ + ≤‫أو‬
2
2 2 3
3
k
π
π π π− ≤ + ≤
‫ﻟﺪﻳﻨﺎ‬2 2 3
3
k
π
π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬
7 8
6 6
k− ≤ ≤
‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k =
‫وﻣﻨﻪ‬
7
3
x
π
=‫أو‬
3
x
π
=‫أو‬
5
3
x
π
= −
‫ﻟﺪﻳﻨﺎ‬‫ﻟﺪﻳﻨﺎ‬
2
2 2 3
3
k
π
π π π− ≤ + ≤‫ﺗﻜﺎﻓﺊ‬
8 7
6 6
k− ≤ ≤
‫ﺗﻜﺎﻓﺊ‬1k = −‫أو‬0k =‫أو‬1k =
‫وﻣﻨﻪ‬
8
3
x
π
=‫أو‬
2
3
x
π
=‫أو‬
4
3
x
π
= −
‫إذن‬
5 4 2 7 8
; ; ; ; ;
3 3 3 3 3 3
S
π π π π π π− − 
=  
 
‫ﺧﻼﺻﺔ‬‫اﻟﻤﻌﺎدﻟﺔ‬sin x a=‫آﺎن‬ ‫إذا‬ ‫ﺣﻼ‬ ‫ﺗﻘﺒﻞ‬ ‫ﻻ‬1 1a a− ∨≺
/ 2 sin 1
2
k x k x x
π
π∈ = + ⇔ ∈ =
/ 2 sin 1
2
k x k x x
π
π∈ = − + ⇔ ∈ = −
‫آﺎن‬ ‫إذا‬1 1a− ≺ ≺‫ﻋﻨﺼﺮ‬ ‫ﻳﻮﺟﺪ‬ ‫ﻓﺎن‬α‫ﻣﻦ‬;
2 2
π π 
−  
‫ﺣﻴﺚ‬sin aα =
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬sin x a=‫ﻓ‬‫ﻲ‬‫هﻲ‬2x kα π= +‫أو‬2x kπ α π= − +‫ﺣﻴﺚ‬k ∈
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬{ } { }2 / 2 /S k k k kα π π α π= + ∈ ∪ − + ∈
‫ﺗﻤﺮﻳﻦ‬‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬( )sin 2 cos 3
3
x x x
π 
∈ + = 
 
] ]
1
;2 sin 2
4 2
x x
π
π π
 
∈ − − = − 
 
‫اﻟﺤﻞ‬--------------------------------------------------------------------------------------
‫ﻧ‬‫ﺤﻞ‬( )sin 2 cos 3
3
x x x
π 
∈ + = 
 
( )sin 2 cos 3
3
x x
π 
+ = 
 
‫ﺗﻜﺎﻓﺊ‬sin 2 sin 3
3 2
x x
π π   
+ = −   
   
‫ﺗﻜﺎﻓﺊ‬2 3 2
3 2
x x k
π π
π+ = − +‫أو‬2 3 2
3 2
x x k
π π
π π+ = − + +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬5 2
6
x k
π
π= +‫أو‬2
6
x k
π
π− = +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬
2
30 5
x k
π
π= +‫أو‬( )2
6
x k
π
π= − + −‫ﺣﻴﺚ‬k ∈
‫إذن‬( )
2
/ 2 /
30 5 6
S k k k k
π π
π π
   
= + ∈ ∪ − + − ∈   
   
‫ﻧﺤﻞ‬] ]
1
;2 sin 2
4 2
x x
π
π π
 
∈ − − = − 
 
1
sin 2
4 2
x
π 
− = − 
 
‫ﺗﻜﺎﻓﺊ‬sin 2 sin
4 6
x
π π   
− = −   
   
‫ﺗﻜﺎﻓﺊ‬2 2
4 6
x k
π π
π− = − +‫أو‬2 2
4 6
x k
π π
π π− = + +‫ﺣﻴﺚ‬k ∈
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5. 5
‫ﺗﻜﺎﻓﺊ‬2 2
12
x k
π
π= +‫أو‬
17
2 2
12
x k
π
π= +‫ﺣﻴﺚ‬k ∈
‫ﺗﻜﺎﻓﺊ‬
24
x k
π
π= +‫أو‬
17
24
x k
π
π= +‫ﺣﻴﺚ‬k ∈
‫أن‬ ‫ﺣﻴﺚ‬ ‫و‬] ];2x π π∈ −‫ﻓﺎن‬
‫أﺟﻞ‬ ‫ﻣﻦ‬
24
x k
π
π= +‫ﻟ‬‫ﺪﻳﻨﺎ‬2
24
k
π
π π π− + ≤≺‫أي‬
25 47
24 24
k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k =
‫إذن‬
23
24 24
x
π π
π= − = −‫أو‬
24
x
π
=‫أو‬
25
24 24
x
π π
π= + =
‫أﺟﻞ‬ ‫ﻣﻦ‬‫ﻟﺪﻳﻨﺎ‬
17
2
24
k
π
π π π− + ≤≺‫وﻣﻨﻪ‬
41 31
24 24
k− ≤≺‫ﻣﻨﻪ‬ ‫و‬1k = −‫أو‬0k =‫أو‬1k =
‫إذن‬
17 7
24 24
x
π π
π= − = −‫أو‬
17
24
x
π
=‫أو‬
17 41
24 24
x
π π
π= + =
‫وﻣﻨﻪ‬
23 7 17 25 41
; ; ; ; ;
24 24 24 24 24 24
S
π π π π π π 
= − − 
 
3-‫اﻟﻤﻌﺎدﻟﺔ‬tan x a=
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan 1x x∈ = −
‫ﻧﻌﺘﺒﺮ‬∆‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫أﺻﻠﻬﺎ‬ ‫ﻓﻲ‬I،
‫اﻟﻨﻘﻄﺔ‬ ‫ﻧﺄﺧﺪ‬T‫ﻣﻦ‬∆‫ﺣﻴﺚ‬1−‫أﻓﺼﻮل‬T‫ﻓﻲ‬‫اﻟﻤﺤﻮر‬∆
‫اﻟﻤﺴﺘﻘﻴﻢ‬( )OT‫ﻳﻘﻄﻊ‬‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬( )C
‫اﻟﻨﻘﻄﺘﻴﻦ‬ ‫ﻓﻲ‬M‫و‬'M‫أن‬ ‫ﻧﻌﻠﻢ‬tan( ) 1
4
π
− = −
‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬
4
π
−‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬M
‫أن‬ ‫وﺑﻤﺎ‬tan( ) tanx k xπ+ =‫ﻟﻜﻞ‬/
2
x k k
π
π
 
∈ − + ∈ 
 
‫ﻓﺎن‬‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻠﻮل‬‫هﻲ‬/
4
x k k
π
π
−
= + ∈
‫اذن‬/
4
S k k
π
π
− 
= + ∈ 
 
‫ﺧﺎﺻﻴﺔ‬
tan /x a x k kα π= ⇔ = + ∈‫ﺣﻴﺚ‬α‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬tan x a=‫ﻓﻲ‬;
2 2
π π 
−  
‫اﻟﻤﻌﺎدﻟﺘﻴﻦ‬ ‫ﺣﻞ‬ ‫ﺗﻤﺮﻳﻦ‬
[ ]0;3 tan 2 3x xπ∈ =
tan 2 tan
3
x x x
π 
∈ − = − 
 
II-‫اﻟﻤﺘ‬‫اﻟ‬ ‫ﺮاﺟﺤﺎت‬‫ﻤﺜﻠﺜﻴﺔ‬
‫ﻣﺜﺎل‬1
‫ﺣﻞ‬] ]
1
; cos
2
x xπ π∈ − ≥
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬] ]
1
; cos
2
x xπ π∈ − =
‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﺣﻞ‬ ‫ﺧﻄﻮات‬ ‫ﺑﺈﺗﺒﺎع‬
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6. 6
] ]
1
; cos
2
x xπ π∈ − =‫ﺗﻜﺎﻓﺊ‬
3
x
π
=‫أو‬
3
x
π
= −
‫ﻟﺘﻜﻦ‬
3
M
π 
 
 
‫و‬'
3
M
π 
− 
 
‫اﻟﻤﺜﻠﺜﻴﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬
‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬
‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮس‬'M IM 
 
‫ﻓﻲ‬] ];π π−
‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬;
3 3
S
π π− 
=   
‫ﻣﺜﺎل‬2‫ﺣﻞ‬[ [
1
0;3 cos
2
x xπ∈ ≥
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﻻ‬ ‫ﻧﺤﻞ‬[ [
1
0;3 cos
2
x xπ∈ =
[ [
1
0;3 cos
2
x xπ∈ =‫ﺗﻜﺎﻓﺊ‬
3
x
π
=‫أو‬
7
3
x
π
=‫أو‬
5
3
x
π
=
3
π
‫و‬
7
3
π
‫اﻟﻨﻘﻄﺔ‬ ‫ﻟﻨﻔﺲ‬ ‫ﻣﻨﺤﻨﻴﻴﻦ‬ ‫أﻓﺼﻮﻟﻴﻦ‬M،
‫ﻧﻌﺘﺒﺮ‬
5
3
π
‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬'M
‫ﻟﻠﻨﻘﻂ‬ ‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬( )C
‫اﻟﻘﻮس‬ ‫اﻟﻰ‬ ‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬'M IM 
 
‫ﻓﻲ‬[ [0;3π
‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬
5 7
0; ;
3 3 3
S
π π π   
= ∪      
‫ﻣﺜﺎل‬3
‫ﺣﻞ‬[ ]0;2 tan 3x xπ∈ ≥
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻧﺤﻞ‬[ ]0;2 tan 3x xπ∈ =
[ ]0;2 tan 3x xπ∈ =‫ﺗﻜﺎﻓﺊ‬
3
x
π
=‫أو‬
4
3
x
π
=
‫ﻧﻌﺘﺒﺮ‬
3
π
‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬A
‫و‬
4
3
π
‫ﻟﻠﻨﻘﻄﺔ‬ ‫ﻣﻨﺤﻨﻲ‬ ‫أﻓﺼﻮل‬B
‫اﻟﻤﻨﺤﻨﻴﺔ‬ ‫اﻷﻓﺎﺻﻴﻞ‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫هﻲ‬ ‫اﻟﻤﺘﺮاﺟﺤﺔ‬ ‫ﺣﻠﻮل‬ ‫ﻣﺠﻤﻮﻋﺔ‬
‫ﻟﻠﻨﻘﻂ‬( )C‫ﺗﻨﺘﻤﻲ‬ ‫اﻟﺘﻲ‬‫إﻟﻰ‬‫اﻟﻘﻮﺳﻴﻦ‬ ‫اﺗﺤﺎد‬AJ 
 
‫و‬'BJ 
 
‫ﻓﻲ‬[ ]0;2π
‫هﻲ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وهﺬﻩ‬
4 3
; ;
3 2 3 2
S
π π π π   
= ∪      
‫ﺗﻤﺮﻳﻦ‬
‫ﺣﻞ‬] ]
1
; sin
2
x xπ π
−
∈ −
] ]
1
0;4 sin
2
x xπ
−
∈
[ ]0;2 tan 1x xπ∈ ≺
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7. 7
‫ﺣﻠﻬﺎ‬ ‫ﻓﻲ‬ ‫ﺗﺆول‬ ‫ﻣﺘﺮاﺟﺤﺎت‬‫إﻟﻰ‬‫أ‬ ‫ﻣﺘﺮاﺟﺤﺎت‬‫ﺳﺎﺳﻴﺔ‬
‫ﺗﻤﺮﻳﻦ‬
‫ﺣﻞ‬
[ ]
1
; sin
3 2
x x
π
π π
 
∈ − − ≤ 
 
[ ]0; tan3 3x xπ∈
] ] ( )2
; 4cos 2 1 2 cos 2 0x x xπ π∈ − − + + ≤
] ]
1 tan
; 0
sin 2
x
x
x
π π
+
∈ − ≥
III-‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰواﻳﺎ‬–‫اﻟﺪاﺋﺮﻳﺔ‬ ‫اﻟﺮﺑﺎﻋﻴﺎت‬
1-‫ﺗﻌﺮﻳﻒ‬
•‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫اﻟﺪاﺋﺮة‬ ‫ﻣﺮآﺰ‬ ‫رأﺳﻬﺎ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬
•‫اﻟﻤﺤﻴﻄﻴﺔ‬ ‫اﻟﺰاوﻳﺔ‬:‫ﺿﻠﻌﻴﻬﺎ‬ ‫ﺑﻴﻦ‬ ‫وﺗﺤﺼﺮ‬ ‫ﻟﻠﺪاﺋﺮة‬ ‫رأﺳﻬﺎ‬ ‫ﻳﻨﺘﻤﻲ‬ ‫زاوﻳﺔ‬ ‫هﻲ‬‫ﻣﻦ‬ ‫ﻗﻮﺳﺎ‬‫هﺬﻩ‬‫اﻟﺪاﺋﺮة‬
2-‫ﺧﺎﺻﻴ‬‫ﺎت‬
‫ﻧﺸ‬‫ﺎط‬1
‫ﻟﺘﻜﻦ‬( )C‫ﻣﺮآﺰهﺎ‬ ‫داﺋﺮة‬O‫ﻧﻌﺘﺒﺮ‬A‫و‬B‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺘﻴﻦ‬ ‫ﻧﻘﻄﺘﻴﻦ‬( )C‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﺘﻘﺎﺑﻠﺘﻴﻦ‬ ‫ﻏﻴﺮ‬
‫و‬M‫ﻧﻘﻄ‬‫ﺔ‬‫ﻣﻦ‬( )C‫ﺑﺤﻴﺚ‬AOB‫و‬AMB‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬AB 
 
1-‫أن‬ ‫ﺑﻴﻦ‬2AOB AMB=‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬
‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬
‫ﻧﻘﻄﺔ‬ ‫اﻋﺘﺒﺎر‬ ‫ﻳﻤﻜﻦ‬N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫أ‬ ‫ﺑﺎﺳﺘﻌﻤﺎل‬ ‫و‬/‫اﻟﻤﻄﻠﻮب‬ ‫ﺑﻴﻦ‬ ‫ﻣﺮﺗﻴﻦ‬
2-‫ﻧﻌﺘﺒﺮ‬( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C.‫اﻟﺰاوﻳﺔ‬BAT‫اﻻﻟﺰاوﻳﺔ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻣﺤﻴﻄﻴﺔ‬
‫اﻟﻤﺮآﺰﻳﺔ‬AOB
‫أن‬ ‫ﺑﻴﻦ‬2AOB TAB=
‫اﻟﺤﻞ‬----------------------------
1-‫أ‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫اﻟﻤﺜﻠﺚ‬OBM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬O
‫وﻣﻨﻪ‬2BOM BMOπ= −
‫ﺣﻴﺚ‬ ‫و‬BOM AOBπ= −‫ﻷن‬M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫ﻓﺎن‬2AOB BMO=
‫اذن‬2AOB AMB=
‫ب‬/M‫و‬O‫و‬A‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬ ‫ﻏﻴﺮ‬
N‫ﻣﻦ‬( )C‫ﺣﻴﺚ‬N‫و‬O‫و‬M‫ﻣﺴﺘﻘﻴﻤﻴﺔ‬
‫أ‬ ‫ﺣﺴﺐ‬/‫ﻟﺪﻳﻨﺎ‬2NOB NMB=
‫ﻟﺪﻳﻨﺎ‬OAM‫اﻟﺮأس‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬ ‫ﻣﺜﻠﺚ‬O
‫ﻣﻨﻪ‬ ‫و‬2AOM AMOπ= −
‫ﻟﺪﻳﻨﺎ‬( )AOB NOB AOMπ= − +
‫وﻣﻨﻪ‬( )2 2AOB NMB AMOπ π= − + −
( )2AOB AMO NMB= −
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8. 8
‫إذن‬2AOB AMB=
2/‫أن‬ ‫ﺑﻴﻦ‬2AOB TAB=
( )AT‫ﻟﻠﺪاﺋﺮة‬ ‫اﻟﻤﻤﺎس‬( )C‫وﻣﻨﻪ‬
2
OAB BAT
π
= −
‫ﻟﺪﻳﻨﺎ‬OAB‫ﻓﻲ‬ ‫اﻟﺴﺎﻗﻴﻦ‬ ‫ﻣﺘﺴﺎوي‬‫اﻟﺮأس‬O
‫وﻣﻨﻪ‬2OAB OABπ= −
‫ﺑﺎﻟﺘﺎﻟﻲ‬ ‫و‬2
2
OAB BAT
π
π
 
= − − 
 
‫إذن‬2AOB TAB=
‫ﺧﺎﺻﻴﺔ‬1
‫ﻣﺤﻴ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬ ‫ﺿﻌﻒ‬ ‫هﻮ‬ ‫داﺋﺮة‬ ‫ﻓﻲ‬ ‫ﻣﺮآﺰﻳﺔ‬ ‫زاوﻳﺔ‬ ‫ﻗﻴﺎس‬‫ﻧﻔ‬ ‫ﺗﺤﺼﺮ‬ ‫ﻄﻴﺔ‬‫هﺬﻩ‬ ‫ﺗﺤﺼﺮﻩ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻘﻮس‬ ‫ﺲ‬
‫اﻟﻤﺮآﺰﻳﺔ‬ ‫اﻟﺰاوﻳﺔ‬
‫ﻧﺸﺎط‬2
‫ﻟﺘﻜﻦ‬A‫و‬B‫و‬C‫و‬D‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬( )C‫ﻣﺮآﺰهﺎ‬O
‫أن‬ ‫ﺑﻴﻦ‬ABC ADC π+ =‫أو‬ABC ADC=
‫ﺧﺎﺻﻴﺔ‬2
A‫و‬B‫و‬C‫داﺋﺮة‬ ‫ﻣﻦ‬ ‫ﻧﻘﻂ‬ ‫ﺛﻼث‬( )C‫و‬D‫اﻟﻤﺴﺘﻮى‬ ‫ﻣﻦ‬ ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻧﻘﻂ‬
‫ﺗﻜﻮن‬D‫اﻟﺪاﺋﺮة‬ ‫ﻣﻦ‬( )C‫آﺎن‬ ‫إذا‬ ‫ﻓﻘﻂ‬ ‫و‬ ‫إذا‬ABC ADC π+ =‫أو‬ABC ADC=
3-‫ﻣﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫اﻟﺠﻴﺐ‬ ‫ﻋﻼﻗﺎت‬
‫ﻧﺸﺎط‬3
‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬R‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﻴﻄﺔ‬ABC
‫أن‬ ‫ﺑﻴﻦ‬2
sinsin sin
BC AC AB
R
CA B
= = =‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﺤﺎﻻت‬ ‫ﻓﻲ‬
‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A
‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬
‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬
‫اﻟﺠﻮاب‬
‫أ‬/ABC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A
sin sin 1
2
A
π
= =‫وﻣﻨﻪ‬2
sin
BC
BC R
A
= =
sin
2
AC BC
B
BC R
= =‫وﻣﻨﻪ‬2
sin
AC
R
B
=
sin
2
AB AB
C
BC R
= =‫وﻣﻨﻪ‬2
sin
AB
R
C
=
‫إذن‬2
sinsin sin
BC AC AB
R
CA B
= = =
‫ب‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫ﺟﻤﻴﻊ‬ABC‫ﺣﺎدة‬
‫ﻧﻌﺘﺒﺮ‬D‫ﻧﻘﻄﺔ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬C
DBC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬B
‫ﻟﺪﻳﻨﺎ‬D A≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬
sin
2
BC BC
D
DC R
= =‫وﻣﻨﻪ‬2
sin
BC
R
D
=‫إذن‬2
sin
BC
R
A
=
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9. 9
‫ﻟﺪﻳﻨﺎ‬DAC‫ﻓﻲ‬ ‫اﻟﺰاوﻳﺔ‬ ‫ﻗﺎﺋﻢ‬A
‫و‬CDA B≡‫اﻟﻘﻮس‬ ‫ﻧﻔﺲ‬ ‫ﺗﺤﺼﺮان‬ ‫ﻣﺤﻴﻄﻴﺘﺎن‬ ‫زاوﻳﺘﺎن‬
sin
2
AC AC
CDA
DC R
= =‫وﻣﻨﻪ‬sin
2
AC
B
R
=‫إذن‬2
sin
AC
R
B
=
‫ﻧﻘﻄﺔ‬ ‫ﻧﻌﺘﺒﺮ‬ ‫ﺑﺎﻟﻤﺜﻞ‬‫ﻣﻘﺎﺑﻠﺔ‬‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬A‫ﻧﺒﻴﻦ‬ ‫و‬2
sin
AB
R
C
=
‫إذن‬2
sinsin sin
BC AC AB
R
CA B
= = =
‫ج‬/‫اﻟﻤﺜﻠﺚ‬ ‫زواﻳﺎ‬ ‫إﺣﺪى‬ABC‫ﻣﻨﻔﺮﺟﺔ‬
‫ﻟﻨﻔﺘﺮض‬‫أن‬A‫ﻣﻨﻔﺮﺟﺔ‬
‫ﻧﻌﺘﺒﺮ‬D‫ﻣﻊ‬ ‫ﻗﻄﺮﻳﺎ‬ ‫ﻣﻘﺎﺑﻠﺔ‬ ‫ﻧﻘﻄﺔ‬C
ˆA‫و‬ˆD‫وﻣﻦ‬ ‫ﻣﺘﻜﺎﻣﻠﺘﺎن‬sin sinD A=
sin
2
BC BC
D
DC R
= =‫وﻣﻨﻪ‬2
sin
BC
R
D
=‫إذن‬2
sin
BC
R
A
=
‫اﻟﺰاوﻳﺘﺎن‬B‫و‬C‫ﺣﺎدﺗﺎن‬
‫ب‬ ‫ﺣﺴﺐ‬/‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬2
sin
AB
R
C
=‫و‬2
sin
AC
R
B
=
‫إذن‬2
sinsin sin
BC AC AB
R
CA B
= = =
‫ﺧﺎﺻﻴﺔ‬
‫ﻟﻴﻜﻦ‬ABC‫ﻣﺜﻠﺜﺎ‬‫و‬R‫اﻟﻤﺤﻴﻄﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﻪ‬
2
sinsin sin
BC AC AB
R
CA B
= = =
4-‫اﻟﻤﺜﻠﺚ‬ ‫ﻓﻲ‬ ‫ﻋﻼﻗﺎت‬)‫اﻟﻤﺴﺎﺣﺔ‬-‫اﻟﻤﺤﻴﻂ‬(
‫ﻧﺸﺎط‬
‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬H‫ﻟـ‬ ‫اﻟﻌﻤﻮدي‬ ‫اﻟﻤﺴﻘﻂ‬A‫ﻋﻠﻰ‬( )BC‫و‬S‫ﻣﺴﺎﺣﺘﻪ‬
1-‫أن‬ ‫ﺑﻴﻦ‬( )1
sin
2
S BC AC C= × ×
2-‫ﻟﻴﻜﻦ‬r‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬‫ﺑﺎﻟﻤﺜﻠﺚ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ABC‫و‬O‫ﻣﺮآﺰهﺎ‬
‫أ‬/‫ﻣﺴﺎﺣﺔ‬ ‫أﺣﺴﺐ‬AOC‫ﺑﺪﻻﻟﺔ‬r‫و‬AC
‫ب‬/‫أن‬ ‫ﺑﻴﻦ‬
1
2
S p r= ×‫ﺣﻴﺚ‬p‫اﻟﻤﺜﻠﺚ‬ ‫ﻣﺤﻴﻂ‬ABC
‫ﺧﺎﺻﻴﺔ‬
‫ﻟﻴﻜﻦ‬ABC‫و‬ ‫ﻣﺜﻠﺜﺎ‬r‫و‬ ‫ﺑﻪ‬ ‫اﻟﻤﺤﺎﻃﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﺷﻌﺎع‬S‫ﻣﺴﺎﺣﺘﻪ‬p‫ﻣﺤﻴﻄﻪ‬
( )1
sin
2
S BC AC C= × ×
1
2
S p r= ×
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