Question

In figure 1, O is the centre of a 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°
• B. 40°
• C. 50°
• D. 90°

Answer 1

The correct option is C

50°

It is given that $\angle AOB={100}^{\circ }$
$\Delta$ AOB is isosceles becaues
OA = OB = radius
$\therefore \angle OAB=\angle OBA$
$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$ [Angle sum property of triangle]
$\Rightarrow {100}^{\circ }+\angle OAB+\angle OAB={180}^{\circ }$
$\Rightarrow 2\angle OAB={80}^{\circ }$
$\Rightarrow \angle OAB={40}^{\circ }$
Now, $\angle OAT={90}^{\circ }$ [AT is tangent and OA is radius]
Thus, $\angle BAT=\angle OAT-\angle OAB={90}^{\circ }-{40}^{\circ }={50}^{\circ }$
The correct answer is C.