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]

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 sum property) MathematicsRD Sharma (2020, 2021)All

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