Question

ABCD is a square in the given figure, find x,

• A. $\text{15}{\text{0}}^{\text{o}}$
• B. $\text{13}{\text{0}}^{\text{o}}$
• C. $\text{12}{\text{0}}^{\text{o}}$
• D. None of these

Answer 1

The correct option is B $\text{13}{\text{0}}^{\text{o}}$

In $△ADB(AD=AB\because ABCD$) is square

$\angle A={90}^0\because ABCD$ os square.

So,

$\angle ADB=\angle ABD={45}^0$ (I have proved, although it is very basic thing which maight be remembered )

In $△OEB$

$\angle OEB+{85}^0+{45}^0={180}^0$ (Angle sum property for $△S$)

$\angle OEB={50}^0$

So,

$x={180}^0-\angle OEB={130}^0$