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 :
Show Hint
In a right-angled triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$, the sides are always in the ratio $1 : \sqrt{3} : 2$.
Here, the side opposite to $30^\circ$ is $3$ cm (radius).
Thus, the side adjacent to $30^\circ$ (the tangent) must be $3 \times \sqrt{3} = 3\sqrt{3}$ cm, and the hypotenuse $OP$ must be $3 \times 2 = 6$ cm.
Using this standard ratio saves valuable time during exams.
Step 1: Understanding the Question:
This problem is based on the concepts of "Circles" and "Trigonometry".
We are given a circle with a radius of $3$ cm and center $O$.
Two tangents, $PQ$ and $PR$, are drawn from an external point $P$ to the circle.
The angle between these two tangents is $\angle QPR = 60^\circ$.
We need to find the length of each tangent ($PQ$ and $PR$).
Step 2: Key Formula or Approach:
We use two fundamental geometric properties of tangents to a circle:
• Tangent-Radius Orthogonality: A tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $OQ \perp PQ$ and $OR \perp PR$, which means $\angle OQP = \angle ORP = 90^\circ$.
• Symmetry of Tangents: The line segment joining the center of the circle to the external point bisects the angle between the two tangents. Thus, $OP$ bisects $\angle QPR$:
\[ \angle OPQ = \angle OPR = \frac{1}{2} \angle QPR \]
After establishing these relationships, we can apply trigonometric ratios in the resulting right-angled triangle to find the length of the tangent.
Step 3: Detailed Explanation:
• Let the center of the circle be $O$ and the external point be $P$.
• The radius of the circle is given as $OQ = OR = 3$ cm.
• The total angle between the two tangents is $\angle QPR = 60^\circ$.
• Join the center $O$ to the external point $P$ to form a line segment $OP$.
• Since $OP$ is the angle bisector of $\angle QPR$, we have:
\[ \angle OPQ = \frac{\angle QPR}{2} = \frac{60^\circ}{2} = 30^\circ \]
• Now, look at the triangle $\Delta OQP$. Since the radius is perpendicular to the tangent at the point of contact:
\[ \angle OQP = 90^\circ \]
Therefore, $\Delta OQP$ is a right-angled triangle at vertex $Q$.
• In the right-angled triangle $\Delta OQP$, the side opposite to $\angle OPQ$ is the radius $OQ$, and the adjacent side is the tangent segment $PQ$.
• Using the trigonometric tangent ratio:
\[ \tan(\angle OPQ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OQ}{PQ} \]
• Substitute the known values ($\angle OPQ = 30^\circ$ and $OQ = 3$ cm):
\[ \tan(30^\circ) = \frac{3}{PQ} \]
• Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we substitute this value into the equation:
\[ \frac{1}{\sqrt{3}} = \frac{3}{PQ} \]
• Cross-multiply to solve for $PQ$:
\[ PQ = 3\sqrt{3}\text{ cm} \]
• 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 $3\sqrt{3}$ cm, which corresponds to Option (A).
