In the diagram below, quadrilateral ABCD is inscribed in circle P. 110 72 P. What is mZADC? 1) 70° 2) 72° 3) 108° 4) 110°

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
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### Problem 123

**Problem Statement**:  
In the diagram below, quadrilateral \(ABCD\) is inscribed in circle \(P\).

![Quadrilateral inscribed in a circle](image-url)
- Quadrilateral \(ABCD\) is inscribed in circle \(P\).

- Angles:
  - \(\angle DAB = 110^{\circ}\)
  - \(\angle ABC = 72^{\circ}\)

**Question**: What is \( m\angle ADC \)?
1. \( 70^{\circ} \)
2. \( 72^{\circ} \)
3. \( 108^{\circ} \)
4. \( 110^{\circ} \)

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**Explanation**:

In the diagram, quadrilateral \(ABCD\) is inscribed in circle \(P\). The measure of \( m\angle ADC \) can be found using the properties of cyclic quadrilaterals. A property of cyclic quadrilaterals (quadrilaterals inscribed in a circle) is that the sum of each pair of opposite angles is \(180^{\circ}\).

Given:
- \(\angle DAB = 110^{\circ}\)
- \(\angle ABC = 72^{\circ}\), thus leading to \(\angle ADC = m\angle ABC + m\angle DAB\).

Let's consider:
- \(m\angle DAB + m\angle DCB = 180^{\circ}\).

Since \(m\angle DAB\) is \(110^{\circ}\), then:
\[m\angle DCB = 180^{\circ} - m\angle DAB\]
\[m\angle DCB = 180^{\circ} - 110^{\circ}\]
\[m\angle DCB = 70^{\circ}\]

Thus, the correct answer is:

\[ \boxed{70^{\circ}} \]

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Transcribed Image Text:--- ### Problem 123 **Problem Statement**: In the diagram below, quadrilateral \(ABCD\) is inscribed in circle \(P\). ![Quadrilateral inscribed in a circle](image-url) - Quadrilateral \(ABCD\) is inscribed in circle \(P\). - Angles: - \(\angle DAB = 110^{\circ}\) - \(\angle ABC = 72^{\circ}\) **Question**: What is \( m\angle ADC \)? 1. \( 70^{\circ} \) 2. \( 72^{\circ} \) 3. \( 108^{\circ} \) 4. \( 110^{\circ} \) --- **Explanation**: In the diagram, quadrilateral \(ABCD\) is inscribed in circle \(P\). The measure of \( m\angle ADC \) can be found using the properties of cyclic quadrilaterals. A property of cyclic quadrilaterals (quadrilaterals inscribed in a circle) is that the sum of each pair of opposite angles is \(180^{\circ}\). Given: - \(\angle DAB = 110^{\circ}\) - \(\angle ABC = 72^{\circ}\), thus leading to \(\angle ADC = m\angle ABC + m\angle DAB\). Let's consider: - \(m\angle DAB + m\angle DCB = 180^{\circ}\). Since \(m\angle DAB\) is \(110^{\circ}\), then: \[m\angle DCB = 180^{\circ} - m\angle DAB\] \[m\angle DCB = 180^{\circ} - 110^{\circ}\] \[m\angle DCB = 70^{\circ}\] Thus, the correct answer is: \[ \boxed{70^{\circ}} \] ---
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