In the given figure, PQ and PR are tangents to a circle with centre O and radius 3 cm. If \(\angle QPR = 60^{\circ}\), then the length of each tangent is :
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For two tangents drawn from an external point to a circle of radius \( R \) where the angle between the tangents is \( 2\theta \):
The length of each tangent is given directly by the formula:
\[ L = R \cot \theta \]
Here, \( R = 3 \) and \( 2\theta = 60^{\circ} \implies \theta = 30^{\circ} \).
Thus, \( L = 3 \cot 30^{\circ} = 3\sqrt{3}\text{ cm} \). This is a very useful formula to remember.
Step 1: Understanding the Question:
The topic of this question is Geometry (Properties of Circle Tangents) combined with Trigonometry.
We are given a circle with center \( O \) and radius \( r = 3\text{ cm} \).
Two tangents \( PQ \) and \( PR \) are drawn from an external point \( P \).
The angle between these two tangents is \( \angle QPR = 60^{\circ} \). We need to calculate the length of the tangents \( PQ \) and \( PR \).
Step 2: Key Formula or Approach:
- The tangent to a circle is perpendicular to the radius at the point of contact, so \( \angle OQP = 90^{\circ} \).
- The line segment connecting the external point \( P \) to the center \( O \) bisects the angle between the tangents, so \( \angle OPQ = \frac{1}{2}\angle QPR \).
- In the right-angled triangle \( OQP \), we use the trigonometric ratio:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \]
Step 3: Detailed Explanation:
1. Since \( PQ \) is a tangent and \( OQ \) is the radius through the point of contact \( Q \), the angle \( \angle OQP = 90^{\circ} \).
Thus, \( \Delta OQP \) is a right-angled triangle.
2. The line segment \( OP \) bisects the angle \( \angle QPR \):
\[ \angle OPQ = \frac{\angle QPR}{2} = \frac{60^{\circ}}{2} = 30^{\circ} \]
3. In right-angled triangle \( OQP \), considering the angle \( \angle OPQ = 30^{\circ} \):
The side opposite to \( 30^{\circ} \) is the radius \( OQ \).
The side adjacent to \( 30^{\circ} \) is the tangent \( PQ \).
4. Using the definition of the tangent function:
\[ \tan(\angle OPQ) = \frac{OQ}{PQ} \]
5. Substitute the values of \( \angle OPQ = 30^{\circ} \) and \( OQ = 3\text{ cm} \):
\[ \tan 30^{\circ} = \frac{3}{PQ} \]
6. We know that \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \). Substituting this value:
\[ \frac{1}{\sqrt{3}} = \frac{3}{PQ} \]
7. Solving for \( PQ \) by cross-multiplication:
\[ PQ = 3\sqrt{3}\text{ cm} \]
8. Since tangents drawn from an external point to a circle are equal in length, we have:
\[ PR = PQ = 3\sqrt{3}\text{ cm} \]
Step 4: Final Answer:
The length of each tangent is \(3\sqrt{3}\text{ cm}\), which corresponds to option (A).
