SK Y
Gm
7+
06+
o
Qu
4+
Pri
3
2
1
Un
78
Tal
lev
→ 2
7 8 9
1 2 3 4 5 6
-9-8-7 -6 -5 -4 -3 -2
-2
-3
-4-
y= g(x)
-5+

< 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⭐

nlgqer

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