Question

In the given figure, if $\angle AOC+\angle BOD={266}^{\circ }$, find all the four angles.

Answer 1

Given that,

$\angle AOC+\angle BOD={266}^0$

But In the figure

$\angle AOC=\angle BOD$ (Vertically opposite angle)

Then, $\angle AOC=\angle BOD=dfarc{266}^{0}2={133}^0$

Now,

$\angle AOC+\angle AOD={180}^0$

$\angle AOD={180}^0-{133}^0$

$\angle AOD={47}^0$

And,$\angle AOD=\angle BOC={47}^0$

Hence, $\angle AOC={133}^0,\angle AOD={47}^0,\angle BOD={133}^{0}and\angle BOC={47}^0$

This is the answer.