In the figure given below, $\angle AOC$ and $\angle COD$ are right angles. If$\angle AOB={43}^0$, what is $\angle BOD$?

The correct option is D ${137}^0$
$\angle AOC$ and $\angle COD$ are right angles.
Hence, $\angle AOC={90}^0,\angle COD={90}^0$

$\angle AOB={43}^0$
$\angle AOC=\angle AOB+\angle BOC$
${43}^0+\angle BOC={90}^0$
${43}^0+\angle BOC–{43}^0={90}^0–{43}^0$
$\angle BOC={47}^0$

$\angle BOD=\angle BOC+\angle COD$
$={47}^0+{90}^0$
$={137}^0$