Question

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



What is the measure of $x$?

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${75}^{\circ }$
• D. ${105}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

In a square, each angle measures ${90}^{\circ }$ and the diagonals bisects the angles.
Thus, $\angle BCD={90}^{\circ }$
And, the diagonal $AC$ bisects the angle $BCD$.
That is, $\angle DCO=\angle OCX={45}^{\circ }$.
In triangle $OCX$,
$\angle COD$ is an exterior angle.
We know that an exterior angle is equal to the sum of opposite interior angles.
So, $\angle COD=\angle OCX+\angle CXO$
${105}^{\circ }={45}^{\circ }+x^{\circ }$
${105}^{\circ }-{45}^{\circ }=x^{\circ }$
$x^{\circ }={60}^{\circ }$
Thus, the correct answer is option (b).