What is the solution to the equation?
–4q + 8 = –2q – 16
a.
q = –12
b.
q = –4
c.
q = 4
d.
q = 12

< 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

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step-by-step explanatio89