Question

Vreau preoblema 2 va rog! DAU COROANA!

Answer 1

a)
m(BAC) =m(DAC) =30°
m(BAD) = m(BAC) +m(DAC) =30°+30°=60°
b)
In ∆ AOB dr => sin BAO = BO/AB
1/2=3/AB =>AB= 6 cm
P = 4×l=4×6=24 cm
c)
In ∆ AOB dr => tg 30°= BO/AO
1/√3= 3/AO => AO=3√3
BD=BO+OD =3+3=6 cm
AC= AO+OC=3√3+3√3=6√3 cm
A = (d1×d2/2)=( 6×6√3)/2= 18√3 cm²