Question

In the figure, $\angle DBC={58}^{\circ }.BD$ is a diameter of the circle. Calculate $\angle BDC$

Answer 1

In $△BCD;\angle DBC={58}^{\circ }$
$\angle BCD={90}^{\circ }$ (Angle in the semicircle as $BD$ is diameter)
$\therefore \angle DBC+\angle BCD+\angle BDC={180}^{\circ }$
${58}^{\circ }+{90}^{\circ }+\angle BDC={180}^{\circ }$
$\Rightarrow \angle BDC={180}^{\circ }-({90}^{\circ }+{58}^{\circ })$
$={32}^{\circ }$