In the figure, $ A B $ is a diameter of a circle with centre $ O $ and $ A T $ is a tangent. If $ \angle A O Q=58^{\circ} $, find $ \angle A T Q $.
"
Given:
In the figure, \( A B \) is a diameter of a circle with centre \( O \) and \( A T \) is a tangent.
\( \angle A O Q=58^{\circ} \).
To do:
We have to find \( \angle A T Q \).
Solution:
In the given figure,
$AB$ is the diameter, $AT$ is the tangent and $\angle AOQ = 58^o$
Arc $AQ$ subtends $\angle AOQ$ at the centre and $\angle ABQ$ at the remaining part of the circle.
$\angle ABQ = \frac{1}{2} \angle AOQ$
$= \frac{1}{2} \times 58^o$
$= 29^o$
In $\triangle ABT$,
$\angle BAT = 90^o$
$\angle ABT + \angle ATB = 90^o$
$\angle ABT + \angle ATQ = 90^o$
$29^o + \angle ATQ = 90^o$
$\angle ATQ = 90^o- 29^o$
$= 61^o$
Therefore, \( \angle A T Q = 61^o\).
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