NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.4

Question 6

Step-by-Step Construction

1. Draw a circle with centre \(O\) and radius \(4\,\text{cm}\).
2. We want the angle between tangents = \(60^\circ\). Remember: The angle made at the centre by joining the points of contact (\(\angle AOB\)) and the angle between tangents (\(\angle ATB\)) are supplementary. So, \(\angle AOB = 180^\circ - 60^\circ = 120^\circ\).
3. Now, mark two points \(A\) and \(B\) on the circle such that \(\angle AOB = 120^\circ\).
4. At point \(A\), draw a line \(t_A\) perpendicular to radius \(OA\). At point \(B\), draw a line \(t_B\) perpendicular to radius \(OB\). These are tangents to the circle.
5. Extend \(t_A\) and \(t_B\) until they meet at point \(T\). Now, \(AT\) and \(BT\) are the required tangents. The angle between them is \(60^\circ\).

Step-by-Step Justification

1. In triangle \(AOT\): - \(\angle A = 90^\circ\) (because tangent is perpendicular to radius). - \(\angle AOT = \tfrac{1}{2}\angle AOB = \tfrac{1}{2} \times 120^\circ = 60^\circ\).
2. Using cosine rule in right triangle: \[ \cos 60^\circ = \dfrac{OA}{OT} \] Here, \(OA = 4\,\text{cm}\) and \(\cos 60^\circ = \tfrac{1}{2}\).
3. So, \[ \tfrac{1}{2} = \dfrac{4}{OT} \quad \Rightarrow \quad OT = \dfrac{4}{1/2} = 8\,\text{cm}. \]

Therefore, the distance from the centre of the circle to the intersection point of tangents is \(8\,\text{cm}\).