IF ANGLE BETWEEN TWO TANGENTS DRAWN FROM A POINT P TO A CIRCLE OF RADIUS A AND CENTER O IS 60 DEGREE THEN PROVE THAT AP = A UNDER ROOT 3.

We have the following situation-

We know that tangent is always perpendicular to the radius at the point of contact.

So, $\angle$OAP = 90

We know that if 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.

$\mathrm{So}, \angle \mathrm{OPA} = \frac{1}{2}\angle \mathrm{APB} = \frac{1}{2}\times 60° = 30°$

According to the angle sum property of triangle-

$$\begin{gathered} \mathrm{In} ∆\mathrm{AOP}, \\ \angle \mathrm{AOP} + \angle \mathrm{OAP} + \angle \mathrm{OPA} = 180° \\ \Rightarrow \angle \mathrm{AOP} + 90° + 30° = 180° \\ \Rightarrow \angle \mathrm{AOP} = 60° \end{gathered}$$

So, in triangle AOP

Hence proved.