Question

Draw four lines $OA,OB,OC,OD$ in order, $\angle AOB={48}^o$, $\angle COD={34}^o$. $OP$ bisects $\angle AOB$, $OQ$ bisects $\angle COD$. If $OQ$ is perpendicular to $OP$, calculate $\angle BOC$.

• A. ${49}^o$
• B. ${50}^o$
• C. ${51}^o$
• D. ${48}^o$

Answer 1

The correct option is A ${49}^o$

Since $OP$ is the angle bisector of $\angle AOB$, we have $\angle POB=\frac{48}{2}={24}^o$
Similarly, since $OQ$ is the angle bisector of $\angle COD$, we have $\angle COQ=\frac{34}{2}={17}^o$
Now, $\angle POQ=\angle POB+\angle BOC+\angle COQ$
Given, $\angle POQ={90}^o$
$=>\angle POB+\angle BOC+\angle COQ={90}^o$
$=>{24}^o+\angle BOC+{17}^o={90}^o$
$=>\angle BOC={49}^o$