Question 2 (2 points) (05.04)
18x - 6
9x^5
15x + 5
21x^2
O 1) 5x
14
2) 5x° (3x + 1)
14(3x - 1)
3) 14
5x
4) 14(3x - 1)
5x® (3x +

< 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⭐

3030 ur welcome

step-by-step explanation:

answer: check all that apply, where the options?

step-by-step explanation: