Question

Tetraedrul regulat ABCD are muchiile de lungime 2a, a > 0, iar E şi F sunt mijloacele segmentelor BC, respectiv CD. Calculați:
a) cosinusul unghiului format de o față laterală cu planul bazei;
b) sinusul unghiului EAF​

Answer 1

Fie AO inaltimea tetraedului

AO⊥(BCD)

∡((ADC),(BDC))=∡AFB=∡AFO

(ADC)∩(BDC)=DC

AO⊥BF,

BF⊂(BDC)

AF⊥DC

AF⊂(ADC)

OF⊥DC

muchia=2a

BF inaltime in ΔBDC echilateral⇒

$BF=\frac{l\sqrt{3} }{2} =a\sqrt{3}$

OF se afla la o treime de baza

$OF=\frac{1}{3} \times BF=\frac{a\sqrt{3} }{3}$

AF=BF ( ABCD tetraedru)

$cosAFO=\frac{FO}{AF} =\frac{\frac{a\sqrt{3} }{3} }{a\sqrt{3} } =\frac{1}{3}$

• EF linie mijlocie in ΔBCD⇒ EF=a

AE=AF=a√3

Luam ΔAEF isoscel

Fie AM⊥EF, M mijloc

$MF=\frac{a}{2}$

• Aplicam Pitagora

AM²+MF²=AF²

$AM^2=3a^2-\frac{a^2}{4}=\frac{11a^2}{4}\\\\ AM=\frac{a\sqrt{11} }{2}$

Fie EN⊥AF

• Aflam aria ΔAEF in doua moduri

EN×AF=AM×EF

$EN=\frac{a\sqrt{11} }{2} \times a\times \frac{1}{a\sqrt{3} } =\frac{a\sqrt{33} }{6}$

$sinEAF=sinEAN=\frac{EN}{AE} =\frac{\frac{a\sqrt{33} }{6} }{a\sqrt{3} }=\frac{\sqrt{11} }{6}$