Question

In fig.$\angle ABC={69}^0,\angle ACB={31}^0,find\angle BDC.$

Answer 1

If $\angle ABC={69}^o,\angle ACB={31}^o,$ then $\angle BDC=?$
In $△ABC,$
$\angle ABC+\angle ACB+\angle BAC={180}^o$
$69+31+\angle BAC=180$
$100+\angle BAC=180$
$\angle BAC=180-100$
$\angle BAC={80}^o$
$\angle BAC$ and $\angle BDC$ are angles in same segments. These are equal
$\angle BDC=\angle BAC={80}^o$
$\therefore \angle BDC={80}^o$