Question : $O$ is the orthocentre of triangle $ABC$, and if $\angle BOC = 110^\circ$, then $\angle BAC$ will be:
Option 1: $110^\circ$
Option 2: $70^\circ$
Option 3: $100^\circ$
Option 4: $90^\circ$
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Correct Answer: $70^\circ$
Solution :
Given that $O$ is the orthocentre of triangle $ABC$, and $\angle BOC = 110^\circ$
Join OF, OD, and OE.
In quadrilateral $AFOE,$
$\angle AFO =\angle AEO = 90^\circ$
$\angle FOE =\angle BOC = 110^\circ$ (Vertically opposite angle)
The sum of four angles of a quadrilateral is $360^\circ.$
$⇒\angle AFO +\angle AEO +\angle FOE +\angle FAE = 360^\circ$
$ ⇒90^\circ+90^\circ+110^\circ+\angle FAE = 360^\circ$
$⇒\angle FAE = 360^\circ-290^\circ=70^\circ$
$\therefore \angle FAE =\angle BAC=70^\circ$
Hence, the correct answer is $70^\circ$.
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