Răspuns:
Explicație pas cu pas:
m(∡(VA,BC)=90°=m(∡(㏒VA,d), unde d║BC, A∈d
m(∡(MN,BC)=???
Ducem MD║BC, atunci m(∡(MN,BC)=m(∡(MN, MD)).
Cercetam ΔMND. ND linie mijlocie a ΔVAC, deci ND║VA, dar VA⊥BC, deci si ND⊥BC. MD║BC, atunci ND⊥MD si deci ΔMND dreptunghic in D.
ND=VA:2=m:2, MD=BC:2=l : 2
atunci m(∡(MN,BC)=m(∡(MN, MD))=m(∡NMD) aflam prin tangenta.
tg(∡NMD)=ND:MD
a) m=l√3, deci ND=m:2=l√3 :2, MD=l:2, atunci
tg(∡NMD)=ND:MD=(l√3):l=√3. ⇒m(∡NMD)=60°=m(∡(MN,BC).
b) l=m√3, deci ND=m:2, MD=l :2=(m√3):2, atunci
tg(∡NMD)=ND:MD=m:(m√3)=1/ √3 m(∡NMD)=30°=m(∡(MN,BC).