question_answer

In a cyclic quadrilateral ABCD, \[\angle BCD={{130}^{{}^\circ }}\]and passes through the centre of the circle. Then\[\angle \mathbf{ABD}=\]?

A)  \[{{30}^{{}^\circ }}\]                                   

B)  \[{{40}^{{}^\circ }}\]

C)  \[{{50}^{{}^\circ }}\]                                   

D)  \[{{60}^{{}^\circ }}\]

Correct Answer: B

Solution :

(b): The sum of opposite angles of a concyclic quadrilateral is\[{{180}^{{}^\circ }}\]. \[\therefore \]\[\angle BCD+\angle BAD={{180}^{{}^\circ }}\] \[\Rightarrow \]\[{{130}^{{}^\circ }}+\angle BAD={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle BAD={{180}^{{}^\circ }}-{{130}^{{}^\circ }}={{50}^{{}^\circ }}\] The angle in a semi - circle is a right angle. \[\therefore \]\[\angle BDA={{90}^{{}^\circ }}\] \[\therefore \]In \[\Delta ABD\], \[\therefore \] \[\angle ABD={{90}^{{}^\circ }}-{{50}^{{}^\circ }}={{40}^{{}^\circ }}\]