Question

In the given figure $ABCD$ is a cyclic quadrilateral in which $AC$ and $BD$ are its diagonals. If $\angle DBC={55}^o$ and $\angle BAC={47}^o$, find $\angle BCD$

Answer 1

$\angle CAD=\angle DBC={55}^{\circ }$

$\angle DAB=\angle CAD+\angle BAC$

$=55+47={102}^{\circ }$

Similarly,

$\angle DAB+\angle BCD={180}^{\circ }$

$\angle BCD=180-\angle DAB$

$\therefore \angle BCD=180-102={78}^{\circ }$