Me with this its super annoying
i have tried but it wants to be super specific so : (

Step-by-step explanation:

Angle CBO = 90 - 2x because the radius meets the tangent at 90°

Angle OCB = Angle CBO= 90 - 2x because base angles of an isosceles triangle are equal (OB and OC are radii - so they are equal)

Angle DCO = Angle CDO = x because base angles of an isosceles triangle are equal. (OC and OD are radii - so they are equal)

Angle BCD = Angle DCO + Angle OCB

So Angle BCD = x + (90 - 2x)

Angle BCD = 90 - x

therefore, Angle DOB = 2 × (90 - x)

Angle DOB = 180 - 2x because the angle at the centre is twice the angle at the circumference.

Angle ODA and Angle OBA are both 90° as the angle between the radius and the tangent is 90°

hence, you can work out Angle DAB because angles in a quadrilateral add up to 180°.

Angle DAB = y

y = 360 - 90 - 90 - (180 - 2x)

expand brackets

y = 360 - 90 - 90 - 180 + 2x

y = 2x (angles in a quadrilateral add up to 360°)

practice. practice. practice.