question_answer

ABCD is a cyclic quadrilateral and AD is a diameter. If \[\angle \mathbf{DAC}=\mathbf{6}{{\mathbf{5}}^{{}^\circ }}\] then value of \[\angle \mathbf{ABC}\] is

A)  \[{{55}^{{}^\circ }}\]                                   

B)  \[{{35}^{{}^\circ }}\]

C)  \[{{155}^{{}^\circ }}\]       

D)  \[{{125}^{{}^\circ }}\]

Correct Answer: C

Solution :

(c): In \[\Delta ACD\] \[\angle DAC={{65}^{{}^\circ }}~~\angle ACD={{90}^{{}^\circ }}\] \[\angle D={{180}^{{}^\circ }}-{{65}^{{}^\circ }}-{{90}^{{}^\circ }}={{25}^{{}^\circ }}\] \[\therefore \]\[\angle ABC+\angle ADC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle ABC={{180}^{{}^\circ }}-{{25}^{{}^\circ }}={{155}^{{}^\circ }}\]