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

Given:

Two lines $AB$ and $CD$ intersect at $O$.

$\angle AOC + \angle COB + \angle BOD = 270^o$.

To do:

We have to find the measure of $\angle AOC, \angle COB, \angle BOD$ and $\angle DOA$.

Solution:

We know that,

Sum of the angles about a point is $360^o$.

Therefore,

$\angle AOC + \angle COB + \angle BOD + \angle DOA = 360^o$

$270^o + \angle DOA = 360^o$

$\angle DOA = 360^o - 270^o$

$\angle DOA = 90^o$

$\angle BOC = \angle DOA = 90^o$                  (Vertically opposite angles are equal)

$\angle DOA + \angle BOD = 180^o$           (Linear pair)

$90^o + \angle BOD = 180^o$

$\angle BOD = 180^o-90^o$

$\angle BOD = 90^o$

$\angle AOC = \angle BOD = 90^o$                   (Vertically opposite angles)

Hence, $\angle AOC = 90^o, \angle COB = 90^o, \angle BOD = 90^o$ and $\angle DOA = 90^o$.