Question

In the given figure AB is a chord of the circle and AOC is its diameter such that $\angle ACB={50}^{\circ }$. If AT is the tangent to the circle at the point A, then $\angle BAT$ is equal to:

• A. ${65}^{\circ }$
• B. ${60}^{\circ }$
• C. ${50}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

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

$\angle BAC={180}^{\circ }-(\angle ACB+\angle CBA)$ ....Angle sum property of triangle
$={180}^{\circ }-({50}^{\circ }+{90}^{\circ })$
$={40}^{\circ }$
Now, $\angle BAC+\angle BAT={90}^{\circ }$ ....Tangent is perpendicular to radius
$\Rightarrow \angle BAT={50}^{\circ }$

So, option C is correct.