A parabola whose equation is y=(x-5)^2 -2 has a vertex at
A-(-5,-2)
B-(-5,2)
C-(5,-2)
D-(5,2)

I think the answer is B. Hope it helps!

Step-by-step explanation:

< 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⭐

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step-by-step explanation: