In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to
(a) 45°
(b) 60°
(c) 50°
(d) 55°
(c) In figure, AOC is a diameter of the circle. We know that, diameter subtends an angle 90° at the circle.
So, $\angle A B C=90^{\circ}$
In $\triangle A C B, \quad \angle A+\angle B+\angle C=180^{\circ}$ [since, sum of all angles of a triangle is $180^{\circ}$ ]
$\Rightarrow \quad \angle A+90^{\circ}+50^{\circ}=180^{\circ}$
$\Rightarrow$$\angle A+140=180$
$\angle A=180^{\circ}-140^{\circ}=40^{\circ}$
$\angle A$ or $\angle O A B=40^{\circ}$
Now, $A T$ is the tangent to the circle at point $A$. So, $O A$ is perpendicular to $A T$.
$\therefore$ $\angle O A T=90^{\circ}$ [from figure]
$\Rightarrow$ $\angle O A B+\angle B A T=90^{\circ}$
On putting $\angle O A B=40^{\circ}$, we get
$\Rightarrow \quad \angle B A T=90^{\circ}-40^{\circ}=50^{\circ}$
Hence, the value of $\angle B A T$ is $50^{\circ}$.