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