Question

PQ is tangent to the circle with centre O such that OP = 2OQ. m\(\angle\)OPQ is

• A. 15\(^\circ\)
• B. 60\(^\circ\)
• C. 45\(^\circ\)
• D. 30\(^\circ\)

Hint:

In a right-angled triangle, if one of the sides is exactly half the length of the hypotenuse, then the angle opposite to that side is always \(30^\circ\) because \(\sin 30^\circ = \frac{1}{2}\).
This simple geometric shortcut saves time during multiple-choice questions.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a circle with center O and a tangent line PQ touching the circle at point Q.
The distance from the center O to the external point P is twice the radius OQ, i.e., \(OP = 2OQ\).
We need to determine the measure of the angle \(\angle OPQ\).

Step 2: Key Formula or Approach:
1. The radius OQ drawn to the point of contact Q is perpendicular to the tangent PQ:
\[ \angle OQP = 90^\circ \]
This makes \(\triangle OQP\) a right-angled triangle with hypotenuse \(OP\).
2. In the right-angled triangle \(\triangle OQP\), we can define the sine of angle \(\angle OPQ\) as:
\[ \sin \angle OPQ = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{OQ}{OP} \]

Step 3: Detailed Explanation:

• Let us assume the radius of the circle is \(r\). Thus, \(OQ = r\).

• According to the given relation, the length of \(OP\) is:
\[ OP = 2OQ = 2r \]

• Since \(PQ\) is tangent to the circle at \(Q\), the radius \(OQ\) is perpendicular to \(PQ\).
- Therefore, \(\triangle OQP\) is a right-angled triangle at \(Q\).

• Now, write down the trigonometric ratio for sine of angle \(\angle OPQ\):
\[ \sin \angle OPQ = \frac{OQ}{OP} \]

• Substitute the values \(OQ = r\) and \(OP = 2r\) into the ratio:
\[ \sin \angle OPQ = \frac{r}{2r} = \frac{1}{2} \]

• For any acute angle, we know that \(\sin \theta = \frac{1}{2}\) when \(\theta = 30^\circ\).
- Therefore:
\[ m\angle OPQ = 30^\circ \]


Step 4: Final Answer:
The measure of \(\angle OPQ\) is 30\(^\circ\). This matches option (D).