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]

Answer 1

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.

∴ $\angle \mathrm{ABQ}=\frac{1}{2}\angle \mathrm{AOQ}=\frac{1}{2}\times 58°=29°$

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)

$$\begin{gathered} \Rightarrow 90°+29°+\angle \mathrm{ATB}=180°\left(\angle \mathrm{BAT}=\angle \mathrm{OAT}\mathrm{and}\angle \mathrm{ABT}=\angle \mathrm{ABQ}\right) \\ \Rightarrow \angle \mathrm{ATB}=180°-119°=61° \\ \therefore \angle \mathrm{ATQ}=\angle \mathrm{ATB}=61° \end{gathered}$$