A) $ 125{}^\circ $
B) $ 55{}^\circ $
C) $ 150{}^\circ $
D) $ 110{}^\circ $
Correct Answer: D
$ \therefore $ $ \angle ABC+\angle ACB=180{}^\circ -70{}^\circ =110{}^\circ $ … i) In $ \Delta BCF, $ $ \angle CFB+\angle FBC+\angle FCB=180{}^\circ $
$ \Rightarrow $ $ \angle FBC+\angle FCB=90{}^\circ $ … (ii) Similarly, in $ \Delta BCE $ $ \angle ECB+\angle EBC=90{}^\circ $ … (iii) On adding Eqs, (ii) and (iii), we get $ \angle FBC+\angle FCB+\angle ECB+\angle ECB=180{}^\circ $ $ \angle FCB+\angle EBC=180{}^\circ -110{}^\circ =70{}^\circ $ Now, in $ \Delta BOC $ $ \angle BOC+\angle OBC+\angle OCB=180{}^\circ $ $ \angle BOC=180{}^\circ -70{}^\circ =110{}^\circ $ Alternate Method Given, $ \angle A=70{}^\circ $ $ \angle AFC=\angle AEB=90{}^\circ $ In quadrilateral AFOE, $ \angle FOE+\angle BAC+\angle AFC+\angle AEB=360{}^\circ $ [ $ \because $ The sum of the internal angle of a quadrilateral is equal to $ 360{}^\circ $ ]
$ \Rightarrow $ $ \angle FOE+70{}^\circ +90{}^\circ +90{}^\circ =360{}^\circ $
$ \Rightarrow $ $ \angle FOE+360{}^\circ -250{}^\circ =110{}^\circ $ $ \angle BOC=\angle FOE=110{}^\circ $ [vertically opposite angles]
BETA
