Question

In Fig.1, O is the centre of circle , AB is a chord and AT is the tangent at A. If $\angle AOB={100}^{\circ }$, then $\angle BAT$ is equal to

• A. ${100}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${50}^{\circ }$

$

$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$ (Angle Sum property)

${100}^{\circ }+2.\angle OAB={180}^{\circ }$

$\angle OAB={40}^{\circ }$

Radius OA is the perpendicular to the tangent at point A

$\angle OAT={90}^{\circ }$.

Thus, $\angle BAT=\angle OAT-\angle OAB$

Thus, $\angle BAT={90}^{\circ }-{40}^{\circ }$

$\angle BAT={50}^{\circ }$

Therefore option C is the answer.