Step-by-step explanation:
I have a slightly different way to solve this.
Let P be the mid-point of NO.
As MN = MO, MNO is an isosceles triangle,
line MP is ⊥ to NO
As ∠NMO=90°, triangle MNO is similar to triangles PNM and POM.
MP = NP = PO
MP = NO/2
Let Q be the interaction of BK and NO
BQ = BK - QK
As QK = MP
BQ = BK - MP
= 10 - NO/2
Triangle BNO and BAC are similar
BQ/NO = BK/AC
Substituting BQ by 10 - NO/2
(10 - NO/2) / NO = 10 /30
NO = 3*(10 - NO/2)
NO = 30 - 3*NO/2
5*NO/2 = 30
NO = 12