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< 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⭐

20/27

​​

step-by-step explanation:

1 simplify brackets

-\frac{8}{9}\times \frac{5}{6}−

​9

​

​8

​​ ×

​6

​

​5

​​

2 use this rule: \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}

​b

​

​a

​​ ×

​d

​

​c

​​ =

​bd

​

​ac

​​

-\frac{8\times 5}{9\times 6}−

​9×6

​

​8×5

​​

3 simplify 8\times 58×5 to 4040

-\frac{40}{9\times 6}−

​9×6

​

​40

​​

4 simplify 9\times 69×6 to 5454

-\frac{40}{54}−

​54

​

​40

​​

5 simplify \frac{40}{54}

​54

​

​40

​​   to \frac{20}{27}

​27

​

​20

​​