NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.1

Question 6

Step 1: We are given that \(OT = 4\,\text{cm}\). Point \(T\) lies on the tangent \(AT\). So, \(OT\) is a line from the centre \(O\) to point \(T\).

Step 2: By the property of a tangent to a circle: the radius drawn to the point of tangency is perpendicular to the tangent. That means \(OA \perp AT\). So, triangle \(OTA\) is a right-angled triangle at \(A\).

Step 3: In right triangle \(OTA\), - Hypotenuse = \(OT = 4\,\text{cm}\). - Angle at \(T\) = \(30^\circ\).

Step 4: To find \(OA\) (radius), use the sine function: \[ \sin(30^\circ) = \dfrac{OA}{OT} \] Substituting values: \[ \sin(30^\circ) = \dfrac{OA}{4} \] \[ \dfrac{1}{2} = \dfrac{OA}{4} \] \[ OA = 2\,\text{cm} \]

Step 5: To find \(AT\) (the tangent), use the cosine function: \[ \cos(30^\circ) = \dfrac{AT}{OT} \] Substituting values: \[ \dfrac{\sqrt{3}}{2} = \dfrac{AT}{4} \] \[ AT = 4 \times \dfrac{\sqrt{3}}{2} \] \[ AT = 2\sqrt{3}\,\text{cm} \]

Final Answer: \(AT = 2\sqrt{3}\,\text{cm}\). (Option C)