Răspuns:
Explicație pas cu pas:
In ΔEGF
EF=GF/2 GF=60 mm
cos30 =EG/GF √3/2=EG/60
EG=60√3/2=30√3 mm
In Δ EGH
sin 30 =EG/GH GH=2EG=60√3 mm
T Pitagora
GH²=EG²+EH² EH²=GH²-EG²=3600×3-900×3=3(3600-900)
=3×27×100 EH=90 mm
2
IJ=45 mm ∡K=∡J=45°
KI=IJ=45 mm
KJ=45√2 mm