Question

in the circle below, $overline{qs}$ is a diameter. suppose $moverparen{rs}=106^{circ}$ and $mangle srt = 33^{circ}$. find the following. (a) $mangle qrt=square^{circ}$ (b) $mangle qsr=square^{circ}$

Explanation:

Step1: Recall the inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc.

Step2: Find \(m\angle QRT\)

Since \(\overline{QS}\) is a diameter, the arc \(QTS\) is a semi - circle and its measure is \(180^{\circ}\). The inscribed angle \(\angle QRT\) intercepts arc \(QT\). First, we know that the sum of arcs in a circle is \(360^{\circ}\), and given \(m\overparen{RS}=106^{\circ}\). Let's assume we want to find the arc \(QT\). But if we use the fact that \(\angle SRT\) is an inscribed angle that intercepts arc \(ST\) and \(m\angle SRT = 33^{\circ}\), then \(m\overparen{ST}=2m\angle SRT=66^{\circ}\). Since \(\overline{QS}\) is a diameter (\(m\overparen{QTS} = 180^{\circ}\)), and \(m\overparen{RS}=106^{\circ}\), \(m\overparen{ST}=66^{\circ}\), then the arc \(QT\) has measure \(m\overparen{QT}=180^{\circ}-106^{\circ}-66^{\circ}=8^{\circ}\). The inscribed angle \(\angle QRT\) intercepts arc \(QT\), so \(m\angle QRT=\frac{1}{2}m\overparen{QT}\). Since \(m\overparen{QT} = 8^{\circ}\), \(m\angle QRT = 4^{\circ}\).

Step3: Find \(m\angle QSR\)

The inscribed angle \(\angle QSR\) intercepts arc \(QR\). We know that \(m\overparen{QTS}=180^{\circ}\), \(m\overparen{ST} = 66^{\circ}\), so \(m\overparen{QR}=180^{\circ}-66^{\circ}=114^{\circ}\). By the inscribed - angle theorem, \(m\angle QSR=\frac{1}{2}m\overparen{QR}\). So \(m\angle QSR = 57^{\circ}\).

Answer:

(a) \(m\angle QRT = 4^{\circ}\)
(b) \(m\angle QSR=57^{\circ}\)