9. If two tangents inclined at \(60^\circ\) are drawn to a circle of radius 3 cm, then the length of each tangent is
\(\dfrac{3}{2}\sqrt{3}\) cm
6 cm
3 cm
\(3\sqrt{3}\) cm
Step 1: We are given a circle of radius \(r = 3\,\text{cm}\). Two tangents are drawn from an external point such that the angle between them is \(60^\circ\).
Step 2: At the centre of the circle, the angle between the two radii (drawn to the tangent contact points) will be \(180^\circ - 60^\circ = 120^\circ\). (Because in a quadrilateral formed by the centre, the two tangent points, and the external point, the sum of opposite angles is \(180^\circ\)).
Step 3: Now consider the triangle formed by: - the centre of the circle (O), - one tangent point (P), - and the external point (A). In this triangle, angle at the centre is \(120^\circ\). So the half angle is \(120^\circ / 2 = 60^\circ\). This means the angle at OAP is \(30^\circ\).
Step 4: The distance from the centre O to the external point A can be found using \[ OA = \dfrac{r}{\sin 30^\circ} = \dfrac{3}{1/2} = 6\,\text{cm}. \]
Step 5: The tangent length (AP) can be found by Pythagoras theorem in right triangle OAP: \[ AP = \sqrt{OA^2 - OP^2} \] Here, \(OA = 6\,\text{cm},\ OP = r = 3\,\text{cm}\). So, \[ AP = \sqrt{6^2 - 3^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}\,\text{cm}. \]
Final Answer: Each tangent is \(3\sqrt{3}\,\text{cm}\) long.