Question

1. given circle s, find m∠tuv. a. 32° b. 28° c. 58° d. 36°

Explanation:

Step1: Identify the diameter and semicircle

Since \( ST \) is a diameter (passes through center \( S \)), arc \( T V U \) is a semicircle? Wait, no, arc \( T V \) and arc \( V U \)? Wait, given arc \( U V \) is \( 116^\circ \)? Wait, no, the diagram shows arc \( V U \) (wait, the label is \( 116^\circ \) next to arc from \( U \) to \( V \)? Wait, actually, \( \angle TUV \) is an inscribed angle, and \( ST \) is a diameter, so arc \( TV \) and arc \( VU \) should add up to \( 180^\circ \) (since \( ST \) is diameter, semicircle is \( 180^\circ \)). Wait, the given arc \( U V \) is \( 116^\circ \)? Wait, no, maybe arc \( V U \) is \( 116^\circ \), so arc \( T V \) is \( 180^\circ - 116^\circ = 64^\circ \)? Wait, no, \( \angle TUV \) is an inscribed angle subtended by arc \( T V \). Wait, inscribed angle theorem: measure of inscribed angle is half the measure of its subtended arc. Wait, \( ST \) is diameter, so \( \angle TUV \) is an inscribed angle with \( ST \) as diameter? Wait, no, \( U \) is on the circle, \( T \) and \( V \) are on the circle, \( S \) is center. So arc \( T V \): since \( ST \) is diameter, the semicircle is \( 180^\circ \), so arc \( T V + \) arc \( V U = 180^\circ \). Given arc \( V U = 116^\circ \), so arc \( T V = 180 - 116 = 64^\circ \). Then \( \angle TUV \) is an inscribed angle subtended by arc \( T V \), so \( m\angle TUV = \frac{1}{2} \times \) arc \( T V \). Wait, no: wait, \( \angle TUV \): vertex at \( U \), sides \( U T \) and \( U V \). \( U T \) is a diameter? Wait, \( S \) is center, so \( U T \) is a diameter (since \( S \) is on \( U T \)). So \( \angle TUV \) is an inscribed angle subtended by arc \( T V \). So arc \( T V \) is \( 180 - 116 = 64^\circ \)? Wait, no, that can't be. Wait, maybe I got the arc wrong. Wait, the given \( 116^\circ \) is arc \( U V \), so arc \( T V \) is \( 180 - 116 = 64^\circ \), then \( \angle TUV \) is half of arc \( T V \)? Wait, no, inscribed angle over arc \( T V \): \( \angle TUV \) is inscribed angle, so \( m\angle TUV = \frac{1}{2} \times \) arc \( T V \). Wait, but \( 64^\circ / 2 = 32^\circ \)? No, that's option A? Wait, no, wait, maybe the arc is \( T V \) is \( 64^\circ \), so \( \angle TUV = 32^\circ \)? But wait, let's re-examine. Wait, \( ST \) is diameter, so \( \angle T V U \) would be right angle? No, \( \angle T U V \): vertex at \( U \), so the arc opposite is \( T V \). Wait, maybe the given arc is \( U V \) is \( 116^\circ \), so the arc \( T V \) is \( 180 - 116 = 64^\circ \), then \( \angle TUV \) is half of \( 64^\circ \), which is \( 32^\circ \), which is option A. Wait, but let's check again.

Wait, the problem: circle \( S \), \( ST \) is diameter (since \( S \) is center, \( ST \) passes through \( S \)), so arc \( T V U \) is a semicircle? No, \( ST \) is diameter, so arc \( T V \) and arc \( V U \) make up the semicircle? Wait, no, a diameter divides the circle into two semicircles (each \( 180^\circ \)). So arc \( T V \) + arc \( V U = 180^\circ \). Given arc \( V U = 116^\circ \), so arc \( T V = 180 - 116 = 64^\circ \). Then \( \angle TUV \) is an inscribed angle that intercepts arc \( T V \). By inscribed angle theorem, \( m\angle TUV = \frac{1}{2} \times \) measure of arc \( T V \). So \( \frac{1}{2} \times 64^\circ = 32^\circ \). So the answer is A.

Step2: Verify the inscribed angle theorem

Inscribed angle theorem states that an angle \( \theta \) subtended by an arc at the circumference is half the central angle subtended by the same arc. Here, arc \( T V \) has measure \( 64^\circ \), so the inscribed angle \( \angle TUV \) (subtended by arc \( T V \)) is \( \frac{1}{2} \times 64^\circ = 32^\circ \), which matches option A.

Answer:

A. \( 32^\circ \)