Question

what is the measure of arc qsr?

70°

Explanation:

Step1: Recall the property of tangents and radii

The radius is perpendicular to the tangent at the point of tangency. So, \(\angle OQR = 90^{\circ}\) and \(\angle ORQ=90^{\circ}\) (assuming \(O\) is the center of the circle).

Step2: Use the sum of angles in a quadrilateral

The sum of angles in a quadrilateral is \(360^{\circ}\). Let the central angle \(\angle QOR = x\). We know two angles are \(90^{\circ}\) each (from the radius - tangent property) and the given angle at the external point is \(70^{\circ}\). So, \(90^{\circ}+90^{\circ}+70^{\circ}+x = 360^{\circ}\).

$$x=360^{\circ}-(90^{\circ}+90^{\circ}+70^{\circ})=110^{\circ}$$

Step3: Find the measure of arc \(QSR\)

The measure of a major arc \(QSR\) is \(360^{\circ}-\) measure of minor arc \(QR\). Since the measure of an arc is equal to the measure of its central angle, and we found \(\angle QOR = 110^{\circ}\) (measure of minor arc \(QR\)). Then the measure of arc \(QSR\) is \(360^{\circ}-110^{\circ}\)

$$360^{\circ}-110^{\circ}=250^{\circ}$$

Answer:

\(250\)