Question

If the angle between two tangents drawn from an external point P to a circle of radius' a ' and centre O, the ${60}^o$ the find the length of OP.

Answer 1

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

So, $\angle OAP={90}^0$

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.

So $\angle OPA=\frac{1}{2}\angle APB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

According to the angle sum property of triangle-

In $△AOP+\angle OAP+\angle OPA={180}^{\circ }$

$\to \angle AOP+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$

$\to \angle AOP={60}^{\circ }$



So, $△AOP$

$tan\angle AOP=\frac{AP}{OA}$

$\sqrt{3}=\frac{AP}{a}$

therefore, AP = $\sqrt{3}\times a$