question_answer

In the (not drawn to scale) given figure below, if \[\mathbf{AD}=\mathbf{DC}=\mathbf{BC}\] and \[\angle \mathbf{BCE}=\mathbf{8}{{\mathbf{4}}^{{}^\circ }}\]then \[\angle \mathbf{DBC}\]is:                                                        

A)  \[{{64}^{{}^\circ }}\]                          

B)  \[{{84}^{{}^\circ }}\]

C)  \[{{56}^{{}^\circ }}\]  

D)  \[{{96}^{{}^\circ }}\]

Correct Answer: C

Solution :

(c): Let \[\angle ACD=a=\angle DAC\] \[\therefore \] \[\angle CDB=2a=\angle CBD\] The angles of the base of an isosceles triangle are equal. \[\therefore \] \[\angle ACB={{180}^{{}^\circ }}-{{84}^{{}^\circ }}={{96}^{{}^\circ }}\] \[\Rightarrow \]\[\angle ACD+\angle DCB={{96}^{{}^\circ }}\] \[\Rightarrow \]\[a+{{180}^{{}^\circ }}-4a={{96}^{{}^\circ }}\] \[\Rightarrow \]\[{{180}^{{}^\circ }}-3a={{96}^{{}^\circ }}\]             \[\Rightarrow \]\[3a={{180}^{{}^\circ }}-{{96}^{{}^\circ }}={{84}^{{}^\circ }}\] \[\Rightarrow \] \[a=\frac{{{84}^{{}^\circ }}}{3}={{28}^{{}^\circ }}\] \[\Rightarrow \]\[\angle DBC=2a={{56}^{{}^\circ }}\]