(Csc x)(cot x)+(sin x)

(Hypotenuse/opposite)(Adjacent/opposite)+(Opposite/hypotenuse)

(opposite^2/(Hypotenuse*Adjacent))+(Opposite/hypotenuse)

33

step-by-step explanation: g=6

< 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⭐