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. $5\sqrt{3} \, \text{cm}$
• D. $\frac{5}{\sqrt{3}} \, \text{cm}$

Answer 1

The Correct Option is D

Step 1: Understanding the problem:
We are given two tangents to a circle with a radius of 5 cm, and the angle between the two tangents is 60º. We need to find the length of each tangent.

Step 2: Using geometry to solve the problem:
Let \( O \) be the center of the circle, and let the point of contact of the tangents be \( P \). We know that the radius of the circle is perpendicular to the tangent at the point of contact.
Therefore, the two tangents form an isosceles triangle with the line segment \( OP \) (the radius) being perpendicular to both tangents.

Step 3: Geometry of the triangle:
The angle between the two tangents is 60º, so the angle between the radius \( OP \) and each tangent is 30º (because the angles are symmetric).
Now, we have a right-angled triangle where:
- \( OP = 5 \) cm (radius of the circle),
- The angle between the radius and the tangent is 30º.

Step 4: Using trigonometry to find the length of the tangent:
Let \( L \) be the length of the tangent. In the right-angled triangle, the tangent forms an opposite side to the 30º angle. Using the trigonometric ratio for tangent, we have:
\[ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{L}{5} \] We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{L}{5} \] Solving for \( L \):
\[ L = \frac{5}{\sqrt{3}}\]

Step 5: Conclusion:
The length of each tangent is \[ L = \frac{5}{\sqrt{3}}\]