Question

In the given figure, $ABCD$ is a square. A line sigment $DX$ cuts the side $BC$ at $X$ and the diagonal $AC$ at $O$ such that $\angle COD={105}^o$ and $\angle OXC=x$. The value of $x$ is

• A. ${40}^o$
• B. ${60}^o$
• C. ${80}^o$
• D. ${85}^o$

Answer 1

Answer: (B)

${60}^o$

Since $DX$ is a line segment, then

$\angle COD+\angle COX=180$ {Angle sum property}

$⟹\angle COX=180-105=75-\left(1\right)$

Since $ABCD$ is a square and $AC$ is a diagonal ,

therefore all the angles in a square is ${90}^{\circ }$ each and $AC$ bisects the angle in half. {property}

$⟹\angle ACB\left(\angle OCX\right)=\frac{90}{2}={45}^{\circ }-\left(2\right)$

In traingle $COX$

$⟹\angle OCX+\angle COX+x=180,$ { sum of angles in a triangle is ${180}^{\circ }$)

$⟹x=180-(45+75)$ {using eq 1 and 2}

$⟹x={60}^{\circ }$