Question

In the figure, AB is a chord of the circle and AOC is its diameter such that $\angle ACB={50}^0$. If AT is the tangent to the circle at the point A, then $\angle BAT$ is equal to

• A. ${65}^0$
• B. ${60}^0$
• C. ${50}^0$
• D. ${40}^0$

Answer 1

Answer: (C)

${50}^0$

The line AC passes through O. So it is a diameter of the circle.

$\therefore \angle ABC$ is an angle inscribed in a semicircle.

$\therefore \angle ABC={90}^O.$

Now $\angle ABC+\angle ACB+\angle BAC={180}^O$ ...(sum of the { angles of a $\Delta ={180}^O)$

$\Rightarrow {50}^O+{90}^O+\angle BAC={180}^O$

$\Rightarrow \angle BAC={40}^O$

Again, $AT$ is a tangent to the circle at $A.$

$\therefore \angle CAT={90}^O$

$\Rightarrow \angle BAT+\angle CAB={90}^O$

$\Rightarrow \angle BAT+{40}^o={90}^O$

$\Rightarrow \angle BAT={50}^O.$