Question

In the given figure, O is the centre of the circle and AB is a tangent to it at point B. If $\angle BDC={65}^{\circ }$, then find $\angle BAO$

• A. ${80}^{\circ }$
• B. ${40}^{\circ }$
• C. ${120}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is B

${40}^{\circ }$

Given $\angle BDC={65}^{\circ }$ and AB is tangent to circle with centre O.

$\because$ OB is radius
By Theorem - Tangent is perpendicular to the radius through the point of contact.

$\therefore OB\perp AB$
$\Rightarrow \angle DBC={90}^{\circ }$

In $\Delta BDC$,

$\angle DBC+\angle BDC+\angle BCD={180}^{\circ }$

${90}^{\circ }+{65}^{\circ }+\angle BCD={180}^{\circ }$

$\Rightarrow \angle BCD$ or $\angle OCE$ = ${25}^{\circ }$

$\because OE=OC$ = radius
By Theorem- Angles opposite to equal sides are equal$\Rightarrow \angle OEC=\angle OCE={25}^{\circ }$

Also, $\angle BOE=\angle OEC+\angle OCE$

[Exterior angle = sum of opposite interior angles in a $\Delta$ ]

$\Rightarrow \angle BOE={25}^{\circ }+{25}^{\circ }\Rightarrow \angle BOE={50}^{\circ }$

$\Rightarrow \angle BOA={50}^{\circ }$

In $\Delta AOB$

$\angle AOB+\angle BAO+\angle OBA={180}^{\circ }$

${50}^{\circ }+\angle BAO+{90}^{\circ }={180}^{\circ }\Rightarrow \angle BAO={40}^{\circ }$