Question

$O$ is the orthocentre of $△ABC$. If $\angle BAC={70}^{\circ },$ then $\angle BOC=$

• A. ${140}^{\circ }$
• B. ${110}^{\circ }$
• C. ${100}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B

${110}^{\circ }$

Since, $AD,BE$ and $CF$ are altitudes.

$\Rightarrow AD\perp BC,BE\perp AC$ and $CF\perp AB$

In quadrilateral, $AFOE$:

$\angle AFO={90}^{\circ },\angle AEO={90}^{\circ }$ and $\angle FAE={70}^{\circ }$ [Given]

$\angle AFO+\angle AEO+\angle FAE+\angle FOE={360}^{\circ }$ [Sum of the angles of a quadrilateral is ${360}^{\circ }$]

$\Rightarrow$ ${90}^{\circ }$ + ${90}^{\circ }$ + ${70}^{\circ }$ + $\angle FOE$ = ${360}^{\circ }$

$\Rightarrow \angle FOE$ =${360}^{\circ }$ - ${90}^{\circ }$- ${90}^{\circ }$ - ${70}^{\circ }$ = ${110}^{\circ }$

Now, $\angle FOE$ and $\angle BOC$ are vertically opposite angles.

$\Rightarrow \angle FOE=\angle BOC$

$\Rightarrow \angle BOC={110}^{\circ }$