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Question

# 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. [CBSE 2015]

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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. ∴ $\angle \mathrm{ABQ}=\frac{1}{2}\angle \mathrm{AOQ}=\frac{1}{2}×58°=29°$ Now, AT is the tangent and OA is the radius of the circle through the point of contact A. $\therefore \angle \mathrm{OAT}=90°$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact) In ∆ABT, $\angle \mathrm{BAT}+\angle \mathrm{ABT}+\angle \mathrm{ATB}=180°$ (Angle sum property) $⇒90°+29°+\angle \mathrm{ATB}=180°\left(\angle \mathrm{BAT}=\angle \mathrm{OAT}\mathrm{and}\angle \mathrm{ABT}=\angle \mathrm{ABQ}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{ATB}=180°-119°=61°\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{ATQ}=\angle \mathrm{ATB}=61°$

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