Question

Diagonals of a rectangle $ABCD$ intersect at $O$. If $\angle AOD={70}^o$, then $\angle BCO$ is ___________.

• A. ${70}^o$
• B. ${110}^o$
• C. ${35}^o$
• D. ${55}^o$

Answer 1

The correct option is D ${55}^o$

$\angle AOD=\angle BOC={70}^o$ $\left[Verticallyoppositeangles\right]$

In $△BOC:$

$OB=OC$

$\therefore \angle BCO=\angle CBO$

Applying angle sum property,

$\angle BCO+\angle CBO+\angle BOC={180}^o$

$\Rightarrow \angle BCO+\angle BCO+{70}^o={180}^o$

$\Rightarrow 2\angle BCO={180}^o-{70}^o={110}^o$

$\Rightarrow \angle BCO=\frac{{110}^o}{2}={55}^o$

Hence, the value of $\angle BCO$ is ${55}^o$