Short Note

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If \[\angle\] AOQ = 58º, find \[\angle\] ATQ.

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#### Solution

It is given that \[\angle\] AOQ = 58º.

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

∴\[\angle ABQ = \frac{1}{2}\angle AOQ = \frac{1}{2} \times 58^o = 29^o\]

Now, AT is the tangent and OA is the radius of the circle through the point of contact A.

\[\therefore \angle OAT = 90^o\] (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ∆ABT,

\[\angle BAT + \angle ABT + \angle ATB = 180^o\] (Angle sum property)

\[\Rightarrow 90^o + 29^o + \angle ATB = 180^o \left( \angle BAT = \angle OAT and \angle ABT = \angle ABQ \right)\]

\[ \Rightarrow \angle ATB = 180^o - 119^o = 61^o\]

\[ \therefore \angle ATQ = \angle ATB = 61^o\]

\[ \Rightarrow \angle ATB = 180^o - 119^o = 61^o\]

\[ \therefore \angle ATQ = \angle ATB = 61^o\]

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