Question

if ( overline{np} ) and ( overline{qp} ) are tangent segments, what is the measure of ( angle npq )?
the measurement of ( angle npq=) 。
blank 1:

Explanation:

Step1: Use the property of tangent - radius

Since \(NP\) and \(QP\) are tangent segments to the circle, \(ON\perp NP\) and \(OQ\perp QP\) (a tangent to a circle is perpendicular to the radius at the point of tangency). So, \(\angle ONP = \angle OQP=90^{\circ}\).

Step2: Use the sum of angles in a quadrilateral

The sum of the interior angles of a quadrilateral \(ONPQ\) is \(360^{\circ}\). Let \(\angle NPQ=x\). Then \(\angle NOQ + \angle ONP+\angle NPQ+\angle OQP = 360^{\circ}\). We know that \(\angle ONP = \angle OQP = 90^{\circ}\). Also, \(\angle NOQ=180^{\circ}- 71^{\circ}=109^{\circ}\) (linear - pair of angles).
Substituting the values into the angle - sum formula for the quadrilateral:

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Another way:

Step1: Use the property of tangent segments

Since \(NP\) and \(QP\) are tangent segments from the same external point \(P\) to the circle, \(NP = QP\), and \(\triangle ONP\cong\triangle OQP\) (by the Hypotenuse - Leg congruence criterion in right - triangles, \(ON = OQ\) (radii of the same circle) and \(OP = OP\) (common side)).
Let \(\angle NPO=\angle QPO = y\).
We know that \(\angle ONP = 90^{\circ}\). Also, \(\angle ONQ = 71^{\circ}\), and \(\angle ONP\) and \(\angle NPO\) are related as follows: \(\angle NPO=90^{\circ}-\angle ONP\) (in right - triangle \(ONP\)).
Since \(\angle NPQ = 2y\) (because \(\angle NPO=\angle QPO\)).
We know that the angle between the radius and the tangent is \(90^{\circ}\). The central angle corresponding to the arc \(NQ\) is \(180^{\circ}-71^{\circ} = 109^{\circ}\).
Using the formula for the angle between two tangents \(\angle NPQ=180^{\circ}-2\times71^{\circ}=38^{\circ}\) (the angle between two tangents from an external point \(P\) to a circle is \(180^{\circ}- 2\theta\), where \(\theta\) is the angle between the radius and one of the tangents at the point of tangency. Here, if we consider the non - tangent angle adjacent to the \(71^{\circ}\) angle in the right - triangle formed by radius and tangent, \(\theta = 71^{\circ}\)).

Answer:

\(38^{\circ}\)