question_answer

Inside a \[\square \,ABCD,\]\[\Delta BEC\]is an equilateral triangle. If CE and BD intersect at O, then \[\angle BOC\]is equal to

A) \[60{}^\circ \]

B) \[75{}^\circ \]

C) \[90{}^\circ \]

D) \[120{}^\circ \]

Correct Answer: B

Solution :

[b] \[\angle BEC=60{}^\circ \] (\[\because \,\,\Delta BEC\] is equilateral.) and       \[\angle ECB=60{}^\circ \] and       \[\angle DBC=45{}^\circ \] (\[\because \]BD is diagonal.) Now, in \[\because \]\[\Delta \Beta {\mathrm O}C\] \[\angle BOC+\angle OBC+\angle OCB=180{}^\circ \] \[\Rightarrow \]   \[\angle BOC+45{}^\circ +60{}^\circ =180{}^\circ \] \[\Rightarrow \]   \[\angle BOC=75{}^\circ \]