Help me please I need to pass summer school

3(6) -1

 2(6)+1

Step-by-step explanation:

first find r(6)

r(6) =3(6) -1

      = 18 -1

Then find s(6)

s(6) = 2(6)+1

Then divide

r(6)/s(6) = 3(6) -1

                  2(6)+1

< 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⭐

thats one

step-by-step explanation: