Question

what is the measure of arc qsr? 70° (diagram: circle with points q, r, s, center, two tangents from an external point forming 70° angle)

Explanation:

Step1: Recall tangent - radius property

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

Step2: Use quadrilateral angle sum

The sum of the interior angles of a quadrilateral is \(360^{\circ}\). In quadrilateral \(OQPR\) (let \(P\) be the external point), we know \(\angle QPR = 70^{\circ}\), \(\angle OQR=90^{\circ}\), \(\angle ORQ = 90^{\circ}\). Let \(\angle QOR=x\). Then \(70^{\circ}+90^{\circ}+90^{\circ}+x = 360^{\circ}\). Solving for \(x\), we get \(x=360^{\circ}-(70^{\circ} + 90^{\circ}+90^{\circ})=110^{\circ}\).

Step3: Find the measure of arc QSR

The total measure of a circle is \(360^{\circ}\). Arc \(QR\) (the minor arc) has a central angle of \(110^{\circ}\), so the measure of arc \(QSR\) (the major arc) is \(360^{\circ}- 110^{\circ}=250^{\circ}\).

Answer:

\(250\)