The solution of a set of four linear inequalities has the shape of a parallelogram. One of the inequalities is Y Z 0. What four inequalities could form the system? A. y≥0, x≥0, x + y ≤ 6, x≤2
B. y ≥ 0,y ≥0, y ≥6, x ≤ 2
C. y ≥ 0, x ≥ 0, x ≤ 6, y ≤ 2
D. y ≥ 0, x ≥ 0, x ≤ -6, y ≤ 2

< 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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