Question

In the given figure, O is the centre of a circle and $\angle OAB={50}^{\circ }$. If AD is a diameter, then find $\angle BOD$.

Answer 1

OA = OB (Radii of a circle)
⇒ $\angle$OBA = $\angle$OAB = 50°
In Δ OAB, we have:
$\angle$ OAB + $\angle$OBA + $\angle$AOB = 180° (Angle sum property of a triangle)
⇒ 50° + 50° + $\angle$AOB = 180°
⇒ $\angle$AOB = (180° - 100°) = 80°
Since $\angle$AOB + $\angle$BOD = 180° (Linear pair)
∴ $\angle$BOD = (180° - 80°) = 100°