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: Problem Definition:
A circle with a 5 cm radius has two tangents drawn to it. These tangents intersect at a 60° angle. Determine the length of each tangent.

Step 2: Geometric Properties of Tangents:
Let the circle's center be \( O \), and the points of tangency be \( P \) and \( Q \). Tangents from an external point \( T \) to a circle are equal in length. Let this length be \( l \).
Key properties:
- \( OP = OQ = 5 \, \text{cm} \) (radii).
- The angle between the tangents at \( T \) is 60°.
- The line segment \( OT \) bisects the angle between the tangents, forming two 30° angles. Thus, \( \angle PTO = \angle QTO = 30^\circ \).
- The radius is perpendicular to the tangent at the point of tangency, so \( \angle OPT = \angle OQT = 90^\circ \).
This forms two congruent right-angled triangles, \( OTP \) and \( OTQ \). In triangle \( OTP \), \( OP \) is opposite to \( \angle PTO \).

Step 3: Trigonometric Calculation:
Using the right-angled triangle \( OTP \), we can relate the radius and the tangent length using the tangent function:
\[\tan(\angle PTO) = \frac{OP}{PT}\]
Substituting the known values:
\[\tan(30^\circ) = \frac{5}{l}\]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[\frac{1}{\sqrt{3}} = \frac{5}{l}\]
Solving for \( l \):
\[l = 5\sqrt{3} \, \text{cm}\]
Alternatively, we could first find \( OT \):
\[\sin(\angle PTO) = \frac{OP}{OT}\]
\[\sin(30^\circ) = \frac{5}{OT}\]
\[\frac{1}{2} = \frac{5}{OT} \implies OT = 10 \, \text{cm}\]
Then, using the Pythagorean theorem in triangle \( OTP \):
\[l^2 = OT^2 - OP^2 = 10^2 - 5^2 = 100 - 25 = 75\]
\[l = \sqrt{75} = 5\sqrt{3} \, \text{cm}\]

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