Question

1. in quadrilateral quad, the m∠a = 62°

what is the measure of arc dqu?
118°
236°
62°
124°
clear all

Explanation:

Step1: Recall Inscribed Quadrilateral Property

In a cyclic quadrilateral (quadrilateral inscribed in a circle), an inscribed angle is half the measure of its intercepted arc. Also, the measure of an inscribed angle and the measure of the arc opposite (intercepted by the opposite angle) have a relationship: the measure of the arc intercepted by an angle is twice the angle, and the arc opposite to the angle (the arc that's not intercepted by it) and the angle relate to the total circle (360°) or the fact that opposite angles in a cyclic quadrilateral sum to 180°, but for the arc, the key is that an inscribed angle intercepts an arc, and the arc opposite (the major arc if the angle is acute) can be found by knowing that the measure of an inscribed angle is half the measure of its intercepted arc, and the arc \( DQU \) is intercepted by angle \( \angle A \)? Wait, no—wait, angle \( \angle A \) is an inscribed angle intercepting arc \( DQU \)? Wait, no, actually, in a cyclic quadrilateral, the measure of an inscribed angle is half the measure of its intercepted arc. Wait, angle \( \angle A \) is at vertex \( A \), so the angle \( \angle A \) intercepts arc \( DQU \). Wait, the formula is: the measure of an inscribed angle \( \theta \) is half the measure of its intercepted arc \( s \), so \( \theta = \frac{1}{2} s \), so \( s = 2\theta \). Wait, but wait, angle \( \angle A \) is 62°, so the arc it intercepts (arc \( DQU \)) should be twice that? Wait, no, maybe I got it reversed. Wait, no—wait, the inscribed angle theorem: an angle formed by two chords in a circle with the vertex on the circle (inscribed angle) intercepts an arc, and the measure of the angle is half the measure of the intercepted arc. So if angle \( A \) is 62°, and it intercepts arc \( DQU \), then arc \( DQU \) should be \( 2 \times 62^\circ = 124^\circ \)? Wait, no, wait, maybe the other way. Wait, no, let's think again. In a cyclic quadrilateral, the measure of an angle is half the measure of the arc opposite to it? Wait, no, the intercepted arc. Let's clarify: the angle at \( A \), \( \angle A \), is formed by chords \( AD \) and \( AQ \). So the angle \( \angle A \) intercepts arc \( DQU \). So by the inscribed angle theorem, \( m\angle A = \frac{1}{2} m\overset{\frown}{DQU} \). Wait, no, that would mean \( m\overset{\frown}{DQU} = 2 \times m\angle A \). Wait, \( m\angle A = 62^\circ \), so \( m\overset{\frown}{DQU} = 2 \times 62^\circ = 124^\circ \)? Wait, but let's check the options. The options include 124°, so that must be it. Wait, but let's confirm. The inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So if angle \( A \) is 62°, the arc it intercepts (arc \( DQU \)) is twice that, so 124°. So step 1: Recall inscribed angle theorem: \( m\angle = \frac{1}{2} m\overset{\frown}{\text{intercepted}} \). Step 2: Identify that \( \angle A \) intercepts arc \( DQU \). Step 3: Apply the formula: \( m\overset{\frown}{DQU} = 2 \times m\angle A \). Step 4: Substitute \( m\angle A = 62^\circ \), so \( m\overset{\frown}{DQU} = 2 \times 62^\circ = 124^\circ \).

Step2: Apply the Inscribed Angle Theorem

Given \( m\angle A = 62^\circ \), and \( \angle A \) is an inscribed angle intercepting arc \( DQU \). By the inscribed angle theorem, \( m\angle A = \frac{1}{2} m\overset{\frown}{DQU} \). Solving for \( m\overset{\frown}{DQU} \), we get \( m\overset{\frown}{DQU} = 2 \times m\angle A \).

Step3: Calculate the Arc Measure

Substitute \( m\angle A = 62^\circ \) into the formula: \( m\overset{\frown}{DQU} = 2 \times 62^\c…

Answer:

124° (the option with 124°)