Question

In the adjoining figure, $ABCD$ is a square. A line segment $CX$ cuts $AB$ at $X$ and the diagonal $BD$ at $O$ such that $\angle$ $COD=$ ${80}^0$ and $\angle$ $OXA=$ $x^0$ . Find the value of $x$.

Answer 1

Given, that $\angle COD={80}^o$

$\Rightarrow \angle DOX=180-80={100}^o$

In Triangle $△XOB$,

$\angle XBO={45}^o$

and $\angle XOB=180-\angle DOX=180-100={80}^o$

Therefore, $\angle OXB=180-80-45={55}^o$

From the figure,

$\Rightarrow x=180-\angle OXB=180-55={125}^o$