The square of the sum of two consecutive positive even integers is 4048 more than the sum of the squares of these two numbers. find the two numbers.

44 and 46.

Step-by-step explanation:

Let the 2  numbers be x and x + 2  ( because they are consecutive even numbers). So:

(x + (x + 2))^2 = 4048 + x^2 + (x + 2)^2

(2x + 2)^2  - x^2 - (x + 2)^2 = 4048

4x^2 + 8x + 4  - x^2 - (x^2 + 4x + 4)  = 4048

4x^2 - x^2 - x^2 + 8x - 4x + 4 - 4 = 4048

2x^2 + 4x - 4048 = 0

x^2 + 2x - 2024 = 0

(x - 44)(x + 46) = 0

x = 44 because we are given that it is positive.

Therefore the other number is x + 2 = 46.

< 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⭐