Question

In the figure, $\angle DBC={58}^{\circ }.BD$ is a diameter of the circle. Calculate $\angle BEC$ (in degrees).

Answer 1

In $△BCD;\angle DBC={58}^{\circ }$
Then, $\angle BCD={90}^{\circ }$ ...(Angle in the semicircle as $BD$ is diameter).

$\therefore \angle DBC+\angle BCD+\angle BDC={180}^{\circ }$ ...[Angle sum property]
$\Rightarrow {58}^{\circ }+{90}^{\circ }+\angle BDC={180}^{\circ }$
$\Rightarrow \angle BDC={180}^{\circ }-({90}^{\circ }+{58}^{\circ })$ $={32}^{\circ }$.

Now, $\angle BEC+\angle BDC={180}^{\circ }(\because BECD$ is a cyclic quadrilateral)

$\Rightarrow$ $\angle BEC={180}^{\circ }-\angle BDC$
$\Rightarrow$ $\angle BEC$ $={180}^{\circ }-{32}^{\circ }$ $={148}^{\circ }$.