PQ is tangent to the circle with centre O such that OP = 2OQ. m\(\angle\)OPQ is
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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.
Step 1: Understand the setup. PQ is tangent to the circle at Q (point on the circle), O is the center, and OP = 2OQ. Let OQ = r (radius), so OP = 2r. Step 2: Apply the tangent-radius perpendicularity. Since PQ is a tangent at Q, OQ is perpendicular to PQ: $\angle OQP = 90^\circ$. Step 3: Set up the right triangle. In right triangle OQP: Hypotenuse = OP = 2r, one leg OQ = r. Step 4: Use trigonometry to find angle OPQ. \[ \sin(\angle OPQ) = \frac{OQ}{OP} = \frac{r}{2r} = \frac{1}{2} \] Step 5: Solve for the angle. \[ \angle OPQ = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \] Step 6: Conclusion. The measure of $\angle OPQ$ is $30^\circ$. \[ \boxed{30^\circ} \]
