4x + 5y = 3
kx + 15y = 9 find k

12

Step-by-step explanation:

4x + 5y = 3 (i)

kx + 15y = 9(ii)

Dividing eqn (ii) by 3, we get,

k/3 x + 5y = 3(iii)

Subtracting eqn (iii) by eqn (i), we get,

-4x - k/3 x = 0

or, -(4 - k/3 )x = 0

or, 4 - k/3 = 0/x

or, 4 - k/3 = 0

or, 4 = k/3

or, k = 4*3

:. k = 12 (Ans)

< 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⭐