Question

Daca o piramida triunghiulara regulata VABC are latura bazei egala cu 6 cm si inaltimea VO = 6 cm , atunci sinusul unghiului diedru format de planele (VAB) si (VAC) este ...

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Answer 1

Răspuns:

Explicație pas cu pas:

AB=6, VO=6. ΔABC regulat, deci AB=BO√3 ⇒BO=6/√3=6√3/3=2√3.

Din ΔVOB, ⇒VB²=VO²+BO²=6²+(2√3)²=36+12=48=16·3, deci VB=4√3.

VN apotema in ΔVAC, ON=(1/2)·BO=(1/2)·2√3=√3.

Din ΔVON, ⇒VN²=VO²+ON²=6²+(√3)²=36+3=39. Deci VN=√39.

Trasam BM⊥VA, deci si CM⊥VA. Deoarece  (VAB) ∩ (VAC) =VA, ⇒VA⊥(BMC), atunci ∡( (VAB) , (VAC) )=∡BMC.

Calculand aria ΔVAC in 2 moduri, ⇒AC·VN=VA·CM , ⇒6·√39=4√3·CM, deci $CM=\dfrac{\sqrt{39}*4\sqrt{3}}{6}=\dfrac{\sqrt{3}*\sqrt{13}*2\sqrt{3} }{3} =2\sqrt{13}$

Deci CM=BM=2√13.

In ΔBCM,  $BC^{2}=BM^{2}+CM^{2}-2*BM*CM*cos(<BMC),~~6^{2}=(2\sqrt{13})^{2}+ (2\sqrt{13})^{2}-2*2\sqrt{13}*2\sqrt{13}*cos(<BMC),~~\\cos(<BMC)=\dfrac{(2\sqrt{13})^{2}+(2\sqrt{13})^{2}-36}{2*2\sqrt{13}*2\sqrt{13}}=\dfrac{68}{8*13}=\dfrac{17}{26} \\ sin^{2}(<BMC)+cos^{2}(<BMC)=1~~ sin^{2}(<BMC)=1-cos^{2}(<BMC)=1-(\dfrac{17}{26})^{2}=\dfrac{26^{2}-17^{2}}{26^{2}}=\dfrac{9*43}{26^{2}},~~deci~ sin(<BMC)=\dfrac{3\sqrt{43} }{26}$

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