Question

Two lines AB and CD intersect at O. If $\angle AOC+\angle COB+\angle BOD={270}^0$, find the measure of $\angle AOC,\angle COB,\angle BODand\angle DOA$.

Answer 1

Two lines AB and CD intersect at O and $\angle AOC+\angle COB+\angle BOD={270}^0$, But $\angle AOC+\angle COB+\angle BOD+\angle DOA={360}^0$ (Angles at a point)
$\therefore {270}^0+\angle DOA$
$\Rightarrow \angle DOA={360}^0-{270}^0={90}^0$
But $\angle DOA=\angle BOC$
(Vertically opposite angles)
$\therefore \angle BOC={90}^0$
But $\angle DOA+\angle BOD={180}^0$ (Linear pair)
$\Rightarrow {90}^0+\angle BOD={180}^0$
$\therefore \angle BOD={180}^0-{90}^0={90}^0$
But $\angle BOD=\angle AOC$
(Vertically opposite angles)
$\therefore \angle AOC={90}^0$
Hence $\angle AOC={90}^0$,
$\angle COB={90}^0$
$\angle BOD={90}^0$,
and $\angle DOA={90}^0$