IN ΔABD sin B=AD/AB
sin 60=√3/2=12√3/AB⇒
AB=24
inΔABD ,∧BAD=30 inseamna ca BD=AB/2=12
daca ∧DCA=45 si ∧ADC=90⇒∧DAC=45⇒ΔADC isoscel si deci AD=DC=12√3
BC=BD+DC=12+12√3
sin C=sin 45=√2/2=AD/AC
√2/2=12√3/AC
AC=12√6
b)A=( BC×AD)/2={12×(1+√3)×12√3}/2=6(3+√3)