Question

In the given figure, straight lines AB and CD intersect at O. If $\angle AOC+\angle BOD={130}^{\circ }$ then $\angle AOD=?$

$\left(a\right){65}^{\circ }$

$\left(b\right){115}^{\circ }$

$\left(c\right){110}^{\circ }$

$\left(d\right){125}^{\circ }$

Answer 1

ANSWER:
$\left(b\right){115}^{\circ }$
We have :
$\angle AOC=\angle BOD$ Vertically-
Opposite Angles ∴ $\angle AOC+\angle BOD=130\Rightarrow \angle AOC+\angle AOC={130}^{\circ }\because \angle AOC=\angle BOD\Rightarrow 2\angle AOC={130}^{\circ }\Rightarrow \angle AOC={65}^{\circ }$
Now,
$\angle AOC+\angle AOD={180}^{\circ }\because CODisastraightline\Rightarrow {65}^{\circ }+\angle AOD={180}^{\circ }\Rightarrow \angle AOC={115}^{\circ }$