Question

Two lines AB and CD intersect at O. If $\angle AOC+\angle COB+\angle BOD={270}^0$, then $\angle AOC$ =

Answer 1

Two lines AB and. CD intersect at O

$\angle AOC+\angle COB+\angle BOD={270}^0$ ...(i)
But $\angle AOC+\angle COB+\angle BOD+\angle DOA={360}^0$...(ii) Subtracting (i) from (ii),
$\angle DOA={360}^0-{270}^0={90}^0$
But $\angle DOA+\angle AOC={180}^0$
$\therefore \angle AOC={180}^0-{90}^0={90}^0$