question_answer

In \[\Delta ABC\] draw \[BE\bot AC\]and \[CF\bot AB\] and the perpendicular BE and CF intersect at the point O. If \[\angle BAC=70{}^\circ ,\] then the value of \[\angle BOC\] is                                                                                                                  [SSC (CGL) 2012]

A) \[125{}^\circ \]            

B) \[55{}^\circ \]  

C) \[150{}^\circ \]            

D) \[110{}^\circ \]

Correct Answer: D

Solution :

In \[\Delta ABC,\] \[\angle BAC=70{}^\circ \]       [given]

\[\angle BAC+\angle ABC+\angle ACB=180{}^\circ \]

[by angle sum property]

\[\therefore \]\[\angle ABC+\angle ACB=180{}^\circ -70{}^\circ =110{}^\circ \]  ... 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]