How do I solve this I've been on this question for 3 weeks!!

c=3√5

Step-by-step explanation:

it appears the vertices of the inscribed quadrilateral,cuts the edges of the square at certain points that form right angles,with C being the hypotenuse.

c=√3²+6²

c=√9+36

c=√45=3√5

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

10+4√2 is the answer. i dont have the time to explain.