question_answer

In the given figure, ABCD is a square in which AO = AX. What is \[\angle XOB?\]

A)  \[22.5{}^\circ \]

B)  \[25{}^\circ \]

C)  \[30{}^\circ \]

D)  \[45{}^\circ \]

Correct Answer: A

Solution :

Let \[\angle XOB=\theta \] In \[\Delta OXB,\] \[\angle XOB+\angle OBX+\angle OXB=180{}^\circ \] \[\Rightarrow \]\[\theta +45{}^\circ +\angle OXB=180{}^\circ \] \[\Rightarrow \]\[\angle OXB=180{}^\circ -45{}^\circ -\theta =135{}^\circ -\theta \] Here,     \[\angle OXA+\angle OXB=180{}^\circ \] \[\Rightarrow \]\[\angle OXA+135{}^\circ -\theta =180{}^\circ \] \[\Rightarrow \]\[\angle OXA=45{}^\circ +\theta \] In \[\Delta OXA,\]\[AO=OX\] \[\therefore \]\[\angle OXA=\angle AOX=45{}^\circ +\theta \]           (Given) Since, \[\angle AOX+\angle XOB=90{}^\circ \] \[\Rightarrow \]\[45{}^\circ +\theta +\theta =90{}^\circ \Rightarrow 2\theta =45{}^\circ \Rightarrow \theta =225{}^\circ \]