Question

In 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.
• B.
• C.
• D.

Answer 1

The correct option is C

In figure AOC is a diameter of the circle. We know that, diameter subtends an angle ${90}^{\circ }$ at the circle.
So, $\angle ABC={90}^{\circ }$
In $△ABC,\angle A+\angle B+\angle C={180}^{\circ }$ [Since, sum of all angles of a triangle is ${180}^{\circ }$ ]
$\Rightarrow$ $\angle A+{90}^{\circ }+{50}^{\circ }={180}^{\circ }$
$\Rightarrow$ $\angle A+140=180$
$\Rightarrow$ $\angle A={180}^{\circ }-{140}^{\circ }={40}^{\circ }$
$\angle A$ or $\angle OAB$ = ${40}^{\circ }$
Now, AT is the tangent to the circle at point A. So, OA is perpendicular to AT
$\therefore$ $\angle OAT={90}^{\circ }$
$\Rightarrow$ $\angle OAB+\angle BAT={90}^{\circ }$
On putting $\angle OAB={40}^{\circ }$, we get
$\Rightarrow$ $\angle BAT={90}^{\circ }-{40}^{\circ }={50}^{\circ }$
Hence, the value of $\angle BAT$ is ${50}^{\circ }$.