Question

what is the measure of arc qr? 26° 104° 128° 52°

Explanation:

Step1: Recall the inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. Let the measure of arc \(QR\) be \(x\). The inscribed angle \(\angle QSR\) intercepts arc \(QR\). But first, note that \(QS\) is a diameter. The sum of the measures of arcs in a circle is \(360^{\circ}\), and the measure of a semic - circle is \(180^{\circ}\).

Step2: Use the relationship between the inscribed angle and the arc

We know that the inscribed angle \(\angle QSR = 52^{\circ}\). The central angle corresponding to arc \(QR\) (let's call it \(\angle QTR\)) has a measure equal to the measure of arc \(QR\). Also, if we consider the property that the measure of an inscribed angle \(\theta\) and the central angle \(\alpha\) intercepting the same arc: \(\theta=\frac{1}{2}\alpha\). But another way is to use the fact that the angle subtended by an arc at the center is twice the angle subtended at the circumference.
Since \(QS\) is a diameter, the arc \(QRS\) is a semic - circle (\(180^{\circ}\)). Let the measure of arc \(QR\) be \(m\). The inscribed angle \(\angle QSR\) intercepts arc \(QR\). The central angle theorem: If an inscribed angle \(\angle QSR\) intercepts arc \(QR\), then the measure of the central angle \(\angle QTR\) (where \(T\) is the center) is \(2\angle QSR\). But more simply, using the property of the circle:
The measure of an inscribed angle \(\angle QSR\) and the arc \(QR\). We know that the measure of an inscribed angle \(\angle QSR\) and the arc \(QR\) are related as \(m(\text{arc }QR)=2\times m(\angle QSR)\) (when the inscribed angle is not subtended by a diameter - related arc in a wrong way). Wait, another approach:
Since \(QS\) is a diameter, the arc \(QRS\) is \(180^{\circ}\). Let the measure of arc \(QR\) be \(x\). The inscribed angle \(\angle QSR\) intercepts arc \(QR\). The central angle \(\angle QTR\) (where \(T\) is the center) has \(m(\angle QTR)=x\). Also, if we consider the triangle \(QTR\) (isosceles if \(TQ = TR\) as radii). But using the inscribed - angle formula correctly:
The measure of an inscribed angle \(\angle QSR\) that intercepts arc \(QR\) is given by \(m(\angle QSR)=\frac{1}{2}m(\text{arc }QR)\) (when the angle is formed by two chords in the interior of the circle). Wait, no, correction:
The measure of an inscribed angle \(\angle QSR\) that intercepts arc \(QR\) is \(m(\angle QSR) =\frac{1}{2}(m(\text{arc }QR))\). But actually, if we consider the fact that \(QS\) is a diameter. Let's use the property that the sum of arcs:
Let \(m(\text{arc }QR)=x\). The inscribed angle \(\angle QSR\) intercepts arc \(QR\). The measure of an inscribed angle \(\angle QSR\) is \(52^{\circ}\). The measure of the arc \(QR\) is \(2\times52^{\circ}=104^{\circ}\) (because the measure of an inscribed angle is half the measure of its intercepted arc)

Answer:

\(104^{\circ}\)