Question

If two tangents inclined at an angle of 60º are drawn to a circle of radius 5 cm, then the length of each tangent is :

• A. $\frac{5\sqrt{3}}{2} \, \text{cm}$
• B. 10 cm
• C. $\frac{5}{\sqrt{3}} \, \text{cm}$
• D. $5\sqrt{3} \, \text{cm}$

Answer 1

The Correct Option is D

Step 1: Understand the given problem:
We are given a circle with a radius of 5 cm, and two tangents are drawn to the circle that make an angle of 60° with each other. We need to find the length of each tangent.

Step 2: Use the geometry of the tangents:
Let the center of the circle be \( O \), and the points where the tangents touch the circle be \( P \) and \( Q \). The tangents from an external point to a circle are equal in length. Let the external point where the tangents are drawn be \( T \), and let the length of each tangent be \( l \).
We know the following properties: - The distance from the center \( O \) to the external point \( T \) is \( OT \). - The radius \( OP = OQ = 5 \, \text{cm} \), and the angle between the two tangents is 60°.
The two tangents and the radius form a triangle. The angle between the tangents at the external point \( T \) is 60°. Therefore, the angle between the line joining the center \( O \) and the external point \( T \) is 90° (since the tangents are perpendicular to the radius). This forms an isosceles triangle \( OTP \), where \( OT \) is the hypotenuse, and \( OP \) is the radius (5 cm).

Step 3: Use the trigonometric relationship:
We can use the following relationship derived from the angle between the tangents:
\[ OT = \frac{r}{\sin 30^\circ} \] where \( r = 5 \, \text{cm} \) is the radius of the circle, and \( \sin 30^\circ = \frac{1}{2} \). Therefore:
\[ OT = \frac{5}{\frac{1}{2}} = 10 \, \text{cm} \] Now, use the Pythagorean theorem to find the length of the tangent \( l \):
\[ l = \sqrt{OT^2 - r^2} = \sqrt{10^2 - 5^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \, \text{cm} \]

Step 4: Conclusion:
The length of each tangent is \( \boxed{5\sqrt{3} \, \text{cm}} \).