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In the given figure, $ A B $ is the 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 $.
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Given:

AB is the diameter of the given circle and AT is a tangent.
$\angle AOQ=58^o$.

To do:
We have to find $\angle ATQ$.

Solution:

AB is a diameter. This implies it is a straight line.

$\angle AOQ + \angle BOQ=180^o$

$\angle BOQ=180^o-58^o$

$\angle BOQ=122^o$
 
In triangle BOQ,

$OB=OQ$  (Radii of the circle)

$\angle OBQ = \angle OQB$    (Angles opposite to the equal sides are equal)

$\angle OBQ + \angle OQB + \angle BOQ=180^o$ 
 
$122^o + 2(\angle OBQ)=180^o$ 

$2\angle OBQ=180^o-122^o$

$\angle OBQ=\frac{58^o}{2}$

$\angle OBQ=29^o$
 
In triangle ABT,

$\angle ABT + \angle BAT + \angle BTA=180^o$ 

$29^o+90^o+\angle BTA=180^o$   ($\angle ABT=\angle OBQ$ and $\angle BAT=90^o$  as AT is tangent to the circle)

$\angle ATQ=180^o-119^o$    ($\angle BTA=\angle ATQ$)

$\angle ATQ=61^o$

The measure of $\angle ATQ$ is $61^o$.

Updated on: 10-Oct-2022

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