Ina circle with centre O, AB is a diameter and CD is a chord such that ABCD is a trapezium. If $$\angle$$BAC = $$28^\circ$$, then $$\angle$$CAD is equal to:

As per the given question,

ABCD is a trapezium, So, AB and CD are a parallel lines.

We know that $$\angle BAC=\angle ACD  .....$$(alternate angle)

$$\angle ACB=90^\circ$$ angled formed by diameter and chord in semi circle.

We know that ABCD is a cyclic quadrilateral, because all the vertices of the quadrilateral is on the circumference of the circle.

Hence the sum  of the opposite angle=180

Let $$\angle CAD=x$$

So, $$\angle BAD+\angle BCD=180^\circ$$

$$\Rightarrow 28+x+90+28=180$$

$$\Rightarrow x=180-90-28-28=34^\circ$$

Hence, $$x=34^\circ$$