Question

To draw a pair of tangents to a circle which are inclined to each other at an angle of ${120}^{\circ }$, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Let O be the centre of the circle to which a pair of tangents AB and AC from a point A touch the circle at B and C, respectively.
Here, ∠BAC = ${120}^{\circ }$


To find: ∠BOC
The angle between a tangent to a circle and the radius of the same circle passing through the point of contact is ${90}^{\circ }$.
Therefore, ∠OBA = ${90}^{\circ }$ and ∠OCA = ${90}^{\circ }$.
By angle sum property of quadrilaterals,
∠BAC + ∠BOC + ∠OBA + ∠OCA = ${360}^{\circ }$
$\Rightarrow {120}^{\circ }$ + ∠BOC + ${90}^{\circ }+{90}^{\circ }+{360}^{\circ }$
$\Rightarrow \angle BOC={60}^{\circ }$

Hence, the correct answer is option b.