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] |
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