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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);
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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) AD | | BC m<(BG, AD) = m<(BG, BC) = m<(CBG)
1.a1) m<(BG, AD) = m<(CBG) = 45o
1.a2) m<(BG, AE) = ?
1.a2) AE | | BF m<(BG, AE) = m<(BG, BF) = m<(FBG)
1.a2) m<(BG, AE) = m<(FBG) = 45o
1.b) m<(BG, DC) = ?
1.b) DC | | AB m<(BG, DC) = m<(BG, AB) = m<(ABG)
1.b) m<(BG, DC) = m<(ABG) = 90o
1.c) m<(AH, GD) = ?
1.c) AH | | BG m<(AH, GD) = m<(BG, GD) = m<(BGD)
1.c) m<(AH, GD) = m<(BGD) = 60o
1.d) tg<(BH, DC) = ?
1.d) DC | | AB tg<(BH, DC) = tg<(BH, AB) = tg<(ABH)
1.d) tg<(BH, DC) = tg<(ABH) = AH/AB =
1.e) m<(BH, AF) = ?
1.e) AF | | BI m<(BH, AF) = m<(BH, BI) = m<(HBI)
1.e) BI=a ; BH=a ; HI=a ; => BI2+BH2=HI2 m<(BH, AF) = m<(HBI) = 90o
1.f) m<(AH, FC) = ?
1.f) AH | | BG m<(AH, FC) = m<(BG, FC) = m<(FOG)
1.f) m<(AH, FC) = m<(FOG) = 90o
1.g) sin<(FD, EC) = ?
1.g) sin<(FD, EC) = sin(EOF)
1.g) A(EOF) = EF . OM/2 = sin<(FD,EC)=sin<(EOF)=
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.a) m<(HC, AD) = ?
2.a) AD | | EH m<(HC, AD) = m<(HC, EH) = m<(CHE)
2.a) m<(HC, AD) = m<(CHE) = 90o
2.b) m<(HC, AB) = ?
2.b) AB | | CD m<(HC, AB) = m<(HC, CD) = m<(HCD)
2.b) m<(HC, AB) = m<(DCH) = 45o
2.c) m<(HC, AE) = ?
2.c) AE | | DH m<(HC, AE) = m<(HC, DH) = m<(DHC)
2.c) m<(HC, AE) = m<(DHC) = 45o
2.d) tg<(HB, AE) = ?
2.d) AE | | DH tg<(HB, AE) = tg<(HB, DH) = tg<(DHB)
2.d) DBAD, m(<A) = 90o => BD = a tg<(HB, AE) = tg<(DHB) = BD/DH =
2.e) tg<(HB, AD) = ?
2.e) AD | | BC m<(HB, AD) = m<(HB, BC) = m<(HBC)
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) MN | | BC sin<(VM,BC) = sin<(VM,MN)= sin<(VMN)
3.a) VP2 = VM2 – MP2 => VP = sin<(VM,BC)=sin<(VMP)=VP/VM=
3.b) m<(VC, AB) = ?
3.b) AB | | MQ ; VC | | MR m<(VC, AB) = m<(MR, MQ) = m<(RMQ)
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: • m<(VB, DC); • b) m<(VA, VC); • c) sin<(VM, AC); • d) sin<(VM, VN).