Răspuns:
Explicație pas cu pas:
c) sin(∡(VM,(VAD)))=???
AD⊥VN, AD⊥MN, ⊥AD⊥(VMN).
MV este oblica la planul VAD. Trasam ME⊥VN, E∈VN, deci ME⊥(VAD).
Deci ∡(VM,(VAD))=∡(VM,VE), unde VE=pr(VAD)VM.
sin(∡(VM,(VAD)))=sin(∡MVE)=ME/VM, din ΔVME, dreptunghic in E.
Din ΔVBM, dreptunghic in M, ⇒VM²=VB²-BM²=(6√3)²-(3√2)²=36·3-9·2=9·(4·3-2)=9·10, deci VM=3√10cm.
ME vom afla din Aria(ΔVMN)
Aria(ΔVMN)=(1/2)·MN·VO=(1/2)·VN·ME |·2, deci MN·VO=VN·ME.
VO aflam din ΔVOM, VO²=VM²-OM²=9·10-(3√2)²=9·10-9·2=9·(10-2)=9·8=9·4·2, deci VO=√(9·4·2)=6√2cm
Inlocuim in MN·VO=VN·ME, ⇒6√2·6√2=3√10·ME, deci
[tex]ME=\frac{6\sqrt{2} *6\sqrt{2} }{3\sqrt{10} }=\frac{24}{\sqrt{10} }\\ Deci~sin(<MVE)=\frac{ME}{MV}=\frac{24}{\sqrt{10} }:3\sqrt{10}=\frac{24}{3*10} =\frac{4}{5}[/tex]
ME=(6√2·6√2)/(3√10)=24/√10.
Deci sin(∡MVE)=ME/MV=(24/√10):(3√10)=24/(3·10)=4/5
