Question

In a square ABCD, $\angle ABD=3x=\angle ADB,$ then $\angle BDC$?

• A. $6x^o$
• B. $(90-6x{)}^o$
• C. ${45}^o$
• D. None

Answer 1

The correct option is C ${45}^o$

$\angle ADB+\angle BDC={90}^o$ (square)
$3x+\angle BDC={90}^o,\angle BDC=90-3x$
In $△ABD,3x+3x+90={180}^o$
$6x=90,x=15$
$\therefore \angle BDC=90-3\left(15\right)$
$=90-45=45$