Can someone help me please

  y = (-1/80)x^2 +2x

Step-by-step explanation:

The graph of the equation for the arch will go through the points ...

  (0, 0), (160, 0), (80, 80)

We can start by making use of the zeros to write the function ...

  y = a(x -0)(x -160)

To make the graph go through the point (80, 80), we must have ...

  80 = a(80 -0)(80 -160) = (-80^2)a

  -1/80 = a . . . . . . divide by the coefficient of "a"

So, our equation is ...

  y = (-1/80)x(x -160)

  y = (-1/80)x^2 +2x

< 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⭐