Question

Quadrilateral ABCD is inscribed in OZ such that and mZBZC = 78. Find mZDCA.

Answer 1

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.

$\therefore m\angle BAC = \frac{1}{2} \times m\angle BZC\\\\\therefore m\angle BAC = \frac{1}{2} \times 78°\\\\\therefore m\angle BAC = 39°\\\\\because In\: \square ABCD, \: AB ||DC... (Given) \\\\\therefore \angle DCA \cong \angle BAC.. (Alternate\: \angle 's) \\\\\therefore m\angle DCA=m \angle BAC\\\\\huge \purple {\boxed {\therefore m\angle DCA=39°}}$