fun one - thanx 4 posting!!!
let the interaction of BK and NO be P
and NO be x
triangle NBO n ABC are similar by AAA as NO // AC
so BP/BK=NO/AC=x/30
BP=10*x/30=x/3
PK=height of triangle NMO
cuz' NM=MO n m∠NMO=90°
height of triangle NMO=NO/2=x/2
PK=x/2
area of triangle NBO + area ot trapezoid NOCA = area of triangle ABC
BP*NO/2 + (NO+AC)*PK/2 = BK*AC/2
(x/3)*x/2+(x+30)*x/2 = 10*30/2
simplifying x^2/6+x^2/4+30x/4=150
5x^2+90x-1800=0
solving x=12 or x=-30
as length cannot be negative
NO=12
