Answer:
39°
Step-by-step explanation:
Since, angle formed on the circumference of the circle is half of the angle formed at the centre of the circle.
[tex]\therefore m\angle BAC = \frac{1}{2} \times m\angle BZC\\\\
\therefore m\angle BAC = \frac{1}{2} \times 78°\\\\
\therefore m\angle BAC = 39°\\\\
\because In\: \square ABCD, \: AB ||DC... (Given) \\\\
\therefore \angle DCA \cong \angle BAC.. (Alternate\: \angle 's) \\\\
\therefore m\angle DCA=m \angle BAC\\\\
\huge \purple {\boxed {\therefore m\angle DCA=39°}} [/tex]
