unghiul a doua drepte in spatiu
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Unghiul a doua drepte in spatiu
Probleme by Todor
Alex
Problema 1
Fie cubul ABCDEFGH. Sa se determine: a) m<(BG, AD); m<(BG, AE); b) m<(BG, DC); c) m<(AH, GD); d) tg<(BH, DC); e) m<(BH, AF); f) m<(AH, FC);
g) sin<(FD, EC).
1.a1)
m<(BG, AD) = ?
1.a1)
m<(BG, AD) = m<(BG, BC) = m<(CBG)
AD BC
1.a1)
m<(BG, AD) = m<(CBG) = 45o
1.a2)
m<(BG, AE) = ?
1.a2)
m<(BG, AE) = m<(BG, BF) = m<(FBG)
AE BF
1.a2)
m<(BG, AE) = m<(FBG) = 45o
1.b)
m<(BG, DC) = ?
1.b)
m<(BG, DC) = m<(BG, AB) = m<(ABG)
DC AB
1.b)
m<(BG, DC) = m<(ABG) = 90o
1.c)
m<(AH, GD) = ?
1.c)
m<(AH, GD) = m<(BG, GD) = m<(BGD)
AH BG
1.c)
m<(AH, GD) = m<(BGD) = 60o
1.d)
tg<(BH, DC) = ?
1.d)
tg<(BH, DC) = tg<(BH, AB) = tg<(ABH)
DC AB
1.d)
tg<(BH, DC) = tg<(ABH) = AH/AB =2
1.e)
m<(BH, AF) = ?
1.e)
m<(BH, AF) = m<(BH, BI) = m<(HBI)
AF BI
1.e)
m<(BH, AF) = m<(HBI) = 90o
BI=a ; BH=a ; HI=a ; => BI2+BH2=HI2
2 3 5
1.f)
m<(AH, FC) = ?
1.f)
m<(AH, FC) = m<(BG, FC) = m<(FOG)
AH BG
1.f)
m<(AH, FC) = m<(FOG) = 90o
1.g)
sin<(FD, EC) = ?
1.g)
sin<(FD, EC) = sin(EOF)
sin<(FD,EC)=sin<(EOF)=
322
OFEO
A2 EOF
)(
1.g)
A(EOF) = EF . OM/2 = 4
2a21
22a
a2
Problema 2
Fie paralelipipedul dreptunghic ABCDEFGH cu AB = a, BC = a , AE= a .
Determinati: a) m<(HC, AD); b) m<(HC, AB); c) m<(HC, AE); d) tg<(HB, AE); e) m<(HB, AD).
2
2.a)
m<(HC, AD) = ?
2.a)
m<(HC, AD) = m<(HC, EH) = m<(CHE)
AD EH
2.a)
m<(HC, AD) = m<(CHE) = 90o
2.b)
m<(HC, AB) = ?
2.b)
m<(HC, AB) = m<(HC, CD) = m<(HCD)
AB CD
2.b)
m<(HC, AB) = m<(DCH) = 45o
2.c)
m<(HC, AE) = ?
2.c)
m<(HC, AE) = m<(HC, DH) = m<(DHC)
AE DH
2.c)
m<(HC, AE) = m<(DHC) = 45o
2.d)
tg<(HB, AE) = ?
2.d)
tg<(HB, AE) = tg<(HB, DH) = tg<(DHB)
AE DH
2.d)
tg<(HB, AE) = tg<(DHB) = BD/DH = 3
BAD, m(<A) = 90o => BD = a3
2.e)
tg<(HB, AD) = ?
2.e)
m<(HB, AD) = m<(HB, BC) = m<(HBC)
AD BC
2.e)
m<(HB, AD) = m<(HBC) = 45o
Problema 3
Fie tetraedrul regulat (piramida triunghiulara cu toate fetele triunghiuri echilaterale) VABC este M mijlocul lui AC.
Determina\i: a) sin<(VM, BC); b) m<(VC, AB).
3.a)
sin<(VM, BC) = ?
3.a)
sin<(VM,BC) = sin<(VM,MN)= sin<(VMN)
MN BC
3.a)
sin<(VM,BC)=sin<(VMP)=VP/VM=
VP2 = VM2 – MP2 => VP =
411a
633a
3.b)
m<(VC, AB) = ?
3.b)
m<(VC, AB) = m<(MR, MQ) = m<(RMQ)
AB MQ ; VC MR
3.b)
RQ2 = VQ2 – VR2 => RQ2 = VQ2 – VR2
m<(VC, AB) = m<(RMQ) = 90o
Problema 4
In piramida patrulatera regulata VABCD se stie ca VB = AB = a cm ]i M, N sunt mijloacele lui BC, AD. Determinati:
a) m<(VB, DC); b) b) m<(VA, VC); c) c) sin<(VM, AC); d) d) sin<(VM, VN).
4.a)
m<(VB, DC) = ?
4.a)
m<(VB, DC) = m<(VB,AB)= m<(VBA)
CD AB
4.a)
m<(VB, DC) = m<(VBA) = 60o
VAB = echilateral
4.b)
m<(VA, VC) = ?
4.b)
m<(VA, VC) = m<(AVC)
4.b)
AC2 = AB2 + BC2 => AC2 = VA2 + VC2
m<(VA, VC) = m<(AVC) = 90o
4.c)
sin<(VM, AC) = ?
4.c)
sin<(VM,AC) = sin<(VM,MP)= sin<(VMP)
AC MP
4.c)
sin<(VM,AC) =sin<(VMQ) =VQ/VM= 6
30
4.d)
sin<(VM, VN) = ?
4.d)
sin<(VM, VN) = sin<(MVN)
4.d)
sin<(VM,VN)=sin<(MVN)=
322
VNVM
A2 MVN
)(
Problema 5
In prisma triunghiulara regulata ABCA'B'C' se cunosc AB = 10 cm, AA' = 10cm.
Sa se afle m<(AC', CB').
5.
m<(AC’, CB’) = ?
5.
m<(AC’,CB’) = m<(AC’,C’D) = m<(AC’D)
CB’ C’D
5.
AC’2 = AC2 + CC’2 => AC’ = DC’ =
310 AD2 = BD2 - AB2 => AD=AC’=DC’=
310
=> m<(AC’,CB’) = m<(AC’D) = 60o
Problema 6
Fie ABC, (AB) (AC), m<(A) = 72 ]i EBCD un p`trat astfel [nc@t E (ABC). S` se determine:
a) m<(AB, ED); b) m<(MN, DC), unde M ]i N sunt
mijloacele laturilor AB ]i AC; c) m<(MN,EC).
6.a)
m<(AB, ED) = ?
6.a)
m<(AB,ED) = m<(AB,BC) = m<(ABC)
ED BC
6.a)
m<(AB,ED)=m<(ABC)=(180o-72o)/2= 54o
6.b)
m<(MN, DC) = ?
6.b)
m<(MN,DC) = m<(BC,DC) = m<(BCD)
MN BC
6.b)
m<(MN,DC) = m<(BCD) = 90o
EBCD = p`trat
6.c)
m<(MN, EC) = ?
6.c)
m<(MN,EC) = m<(BC,EC) = m<(BCE)
MN BC
6.c)
m<(MN,EC) = m<(BCE) = 45o
ABCD = p`trat
Problema 7
Fie romburile ABCD si ABFE situate in plane diferite astfel inact m<(ABC) = 50 grd., m<(ABF) = 30 .grd si CF = AD. Determina\i:
a) m<(EF, BC); b) m<(EA,DC);c) m<(FB, AD);d) m,(ED, BC); e) m<(ED, FC); f) m<(AF, DC).
7.a)
m<(EF, BC) = ?
7.a)
m<(EF,BC) = m<(AB,BC)= m<(ABC)=50o
EF
7.b)
m<(EA, DC) = ?
7.b)
m<(EA,DC) = m<(AB,FB)= m<(ABF)
DC F
Aten\ie: m(<(EA, DC) trebuie s` fie m`sur` de unghi ascu\it !
7.b)
m<(EA,DC) = m<(ABF) = 30o
7.c)
m<(FB, AD) = ?
7.c)
m<(FB,AD) = m<(FB,BC)= m<(FBC)
AD C
7.c)
m<(FB,AD) = m<(FBC) = 60o
BC = FB = CF => BCF = echilateral
7.d)
m<(ED, BC) = ?
7.d)
m<(ED,BC) = m<(FC,BC)= m<(FCB)
ED FC
7.d)
m<(ED,BC) = m<(FCB) = 60o
BC = FC = BF => BCF = echilateral
7.e)
m<(ED, FC) = ?
7.e)
ED FC => m<(ED,FC) = 0o
EF DC ]i EF = DC => CDEF = paralelogram
7.f)
m<(AF, DC) = ?
7.f)
m<(AF,DC) = m<(AF,AB)= m<(FAB)
DC AB
7.f)
m<(AF,DC) = m<(FAB) = 75o
AB = BF => m<(FAB) = (180o-m<(ABF))/2
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