F'(x)=x'*sinx+x*(sinx)'=sinx+x*cosx ; f'(π/2)=sin(π/2)+(π/2)*cos(π/2)=1
∠A = ∠B
∠DOB = ∠AOC
Значит, ∆DOB ~ ∆COA - по I признаку.
Из подобия треугольников =>
AO/OB = OC/OD
5/OB = 4/6
30 = 4OB
OB = 7,5.
AC/BD = CO/OD = 4/6 = 2/3
SAOC/SBDO = (AC/BD)² = 4/9.