a)
AC diametru, m∡ABC=180°/2=90°
ducem BD⊥AC ⇒ T∡15° ⇒ BD=AC/4=6 cm
AD=x, DC=24-x
teorema inaltimii BD
BD^2=x(24-x)
x^2-24x+36=0
(x-12)^2 - (√108)^2=0
(x-12-6√3)(x+12+6√3)=0
solutia acceptata, x>R=12
x=AD=12+6√3
teorema catetei AB
AB^2=24(12+6√3)
AB=12√(2+√3) cm
b)
aria AOB=12^2 x sin(150)/2=144 x 1/4=36 cm2
