Question

In the figure, given below, $ABCD$ is a cyclic quadrilateral in which $\angle BAD={75}^o;\angle ABD={58}^o$ and $\angle ADC={77}^o$. Find:
$\angle BDC$

Answer 1

Given $ABCD$ is a cyclic quadrilateral

$⟹$opposite angle are supplementary angles

$⟹\angle ADC+\angle ABC={180}^{\circ }⟹\angle ABC={180}^{\circ }-{77}^{\circ }={103}^{\circ }$

$⟹\angle CBD=\angle ABC-\angle ABD={103}^{\circ }-{58}^{\circ }={45}^{\circ }$

And also $\angle DAB+\angle BCD={180}^{\circ }⟹\angle BCD={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

In $△BCD$

$⟹\angle BCD+\angle BDC+\angle CBD={180}^{\circ }$

$⟹\angle BDC={180}^{\circ }-{45}^{\circ }-{105}^{\circ }={30}^{\circ }$