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 $\angle COD={80}^{\circ }$ and $\angle OXA=x^{\circ }$. Find the value of x.

Answer 1

Since, the angles of a square are bisected by the diagonals.
So, $\angle OBX={45}^0$ as $\angle ABC={90}^{\circ }$ and BD bisects $\angle ABC$

And $\angle BOX=\angle COD={80}^0$
[Vertically opposite angles]
∴​ In $△BOX$ , we have:
$\angle AXO=\angle OBX+\angle BOX$
[Exterior angle of $△BOX$ ]
⇒​ $\angle AXO={45}^0+{80}^0={125}^0$
$\therefore x={125}^0$