Derivatives of sin(logx) from first principle

cos(logx)*(1/x)

Step-by-step explanation:

(sin(logx))'=cos(logx)*logx'=cos(logx)*(1/x)

< 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⭐

7

step-by-step explanation: