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Question
line segment su is a diameter of circle v. what is the measure of arc st? 56° 68° 112° 163°
Step1: Recall the inscribed - angle theorem
The measure of an inscribed angle is half the measure of its intercepted arc. In the circle \(V\), \(\angle TSU\) is an inscribed angle and \(\overset{\frown}{TU}\) is the intercepted arc. Given \(\angle TSU = 34^{\circ}\), so \(m\overset{\frown}{TU}=2\times\angle TSU = 68^{\circ}\).
Step2: Recall the property of a semi - circle
Since \(SU\) is a diameter, the measure of arc \(STU\) is \(180^{\circ}\) (a semi - circle has an arc measure of \(180^{\circ}\)). Let \(m\overset{\frown}{ST}=x\). Then \(x + m\overset{\frown}{TU}=180^{\circ}\).
Step3: Solve for \(m\overset{\frown}{ST}\)
We know \(m\overset{\frown}{TU} = 68^{\circ}\), so \(x=180^{\circ}-m\overset{\frown}{TU}\). Substituting the value of \(m\overset{\frown}{TU}\), we get \(x = 180^{\circ}- 68^{\circ}=112^{\circ}\).
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\(112^{\circ}\)