Can someone show me step by step how to simplify the square root of 3 over 5and the square root of 12

I will write square root as sqrt.

sqrt12=sqrt2*2*3

(Notice I wrote 4 as 2*2)

there are 2 twos so in the square roots you can take it out.

2sqrt3

< 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⭐

correct answer:

b.(19,-2)