Question

In Fig. 2, AB is the diameter of a circle with centre O and AT is a tangent. If $\angle AOQ={58}^{\circ }$, find $\angle ATQ$.

Answer 1

We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
$\therefore \angle AOQ=2\angle ABQ$
$\Rightarrow \angle ABQ=\frac{1}{2}\angle AOQ$
$\Rightarrow \angle ABQ=\frac{1}{2}\times {58}^{\circ }={29}^{\circ }$
or $\angle ABT={29}^{\circ }$
We know that the radius is perpendicular to the tangent at the point of contact
$\therefore \angle OAT={90}^{\circ }$ $(OA\perp AT)$
Or $\angle BAT={90}^{\circ }$
Now, in $\Delta BAT$,
$\angle BAT+\angle ABT+\angle ATB={180}^{\circ }$
$\Rightarrow {90}^{\circ }+\angle {29}^{\circ }+\angle ATB={180}^{\circ }$
$\Rightarrow \angle ATB={61}^{\circ }$
$\angle ATQ={61}^{\circ }$