Question

finding arc measures involving two intersecting tangents
what is the measure of arc qsr?
image of a circle with center, points q, r, s on the circle, and two tangents from an external point forming a 70° angle

Explanation:

Step1: Recall the property of tangents and circles

The measure of an angle formed by two tangents intersecting outside a circle is equal to half the difference of the measures of the intercepted arcs. Also, the total measure of a circle is \(360^\circ\). Let the measure of the minor arc \(QR\) be \(x\) and the measure of arc \(QSR\) (the major arc) be \(y\). We know that the angle formed outside is \(70^\circ\), and the formula for the angle formed by two tangents outside the circle is \(\frac{1}{2}(y - x)=70^\circ\), and also \(x + y=360^\circ\).

Step2: Solve for the major arc

From the angle formula: \(y - x = 140^\circ\). And we have \(x=360^\circ - y\). Substitute \(x\) into the first equation: \(y-(360^\circ - y)=140^\circ\). Simplify: \(y - 360^\circ+y = 140^\circ\), \(2y=140^\circ + 360^\circ=500^\circ\)? Wait, no, that's a mistake. Wait, the angle formed by two tangents outside the circle: the formula is \(\text{Angle}=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Also, the radius is perpendicular to the tangent at the point of contact, so the quadrilateral formed by the two radii and the two tangents has two right angles (since radius is perpendicular to tangent). So the sum of the interior angles of a quadrilateral is \(360^\circ\). So the central angle for arc \(QR\) (let's call it \(\theta\)) and the angle outside (\(70^\circ\)) and the two right angles (\(90^\circ\) each) sum to \(360^\circ\). So \(\theta+70^\circ + 90^\circ+90^\circ = 360^\circ\), so \(\theta=360^\circ-(70^\circ + 90^\circ+90^\circ)=110^\circ\). Then the major arc \(QSR\) is \(360^\circ - 110^\circ = 250^\circ\)? Wait, no, wait. Wait, the angle between the two tangents outside is \(70^\circ\), and the central angle for the minor arc \(QR\) is \(180^\circ - 70^\circ\)? No, better to use the tangent - angle formula. The measure of the angle formed by two tangents outside the circle is half the difference of the intercepted arcs. So if the angle is \(70^\circ\), then \(70^\circ=\frac{1}{2}(\text{arc }QSR-\text{arc }QR)\). And arc \(QSR+\text{arc }QR = 360^\circ\). Let arc \(QR = x\), arc \(QSR = 360 - x\). Then \(70=\frac{1}{2}((360 - x)-x)\), \(70=\frac{1}{2}(360 - 2x)\), \(140 = 360 - 2x\), \(2x=360 - 140 = 220\), \(x = 110^\circ\). Then arc \(QSR=360 - 110=250^\circ\)? Wait, no, wait. Wait, the correct formula: the measure of an angle formed by two tangents drawn from an external point to a circle is equal to half the difference of the measures of the intercepted arcs. The intercepted arcs are the major arc and the minor arc between the two points of tangency. So if the external angle is \(70^\circ\), then \(70^\circ=\frac{1}{2}(\text{major arc}-\text{minor arc})\). Let the minor arc be \(m\) and major arc be \(M\). Then \(M - m=140^\circ\) and \(M + m = 360^\circ\). Solving these two equations: add them: \(2M=500^\circ\)? No, that can't be. Wait, I made a mistake in the formula. Wait, no, the angle formed by two tangents outside the circle: the formula is \(\text{Angle}=\frac{1}{2}(\text{major arc}-\text{minor arc})\). But if the angle is outside, the major arc is the larger one. Wait, let's think about the right angles. The radius is perpendicular to the tangent, so \(\angle OQT=\angle ORT = 90^\circ\) (where \(O\) is the center, \(T\) is the external point). Then in quadrilateral \(OQTR\), the sum of angles is \(360^\circ\). So \(\angle QOR+\angle OQT+\angle QTR+\angle ORT = 360^\circ\). We know \(\angle OQT=\angle ORT = 90^\circ\), \(\angle QTR = 70^\circ\), so \(\angle QOR=360^\circ-(90^\circ + 70^\circ+90^\circ)=110^\circ\). So the minor arc \(QR\) is \(110^\circ\) (since the central angle is equal to the measure of its intercepted arc). Then the major arc \(QSR\) is \(360^\circ - 110^\circ=250^\circ\)? Wait, no, wait. Wait, the angle between the two tangents is \(70^\circ\), and the central angle for the minor arc is \(180 - 70 = 110\)? No, the quadrilateral has two right angles, so the central angle is \(360 - 90 - 90 - 70=110\), so the minor arc is \(110^\circ\), so the major arc is \(360 - 110 = 250^\circ\). Wait, but let's check the formula again. The measure of the angle formed by two tangents outside the circle is \(\frac{1}{2}(\text{major arc}-\text{minor arc})\). So \(70=\frac{1}{2}(M - m)\), and \(M + m = 360\). So \(M - m = 140\), \(M + m = 360\). Adding both equations: \(2M=500\), \(M = 250\), \(m = 110\). Yes, that matches. So the measure of arc \(QSR\) is \(250^\circ\).

Answer:

\(250^\circ\)