In the diagram below, tangent DA and secant DBC are drawn to circle O from external point D. such that arc AC = arc BC. If arc BC = 152°, determine and state mZD. 48 56 90 96
In the diagram below, tangent DA and secant DBC are drawn to circle O from external point D. such that arc AC = arc BC. If arc BC = 152°, determine and state mZD. 48 56 90 96
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![### Geometry Problem on Tangents and Secants
#### Problem Statement
In the diagram below, tangent \( DA \) and secant \( DBC \) are drawn to circle \( O \) from external point \( D \), such that arc \( AC \cong \) arc \( BC \). If arc \( BC = 152^\circ \), determine and state \( m\angle D \).
#### Diagram Explanation
- Circle \( O \) has a center at point \( O \).
- Tangent \( DA \) touches the circle at point \( A \).
- Secant \( DBC \) intersects the circle at points \( B \) and \( C \).
- Arc \( AC \) is congruent to arc \( BC \).
- The measure of arc \( BC \) is \( 152^\circ \).
#### Multiple Choice Options
- \( 48^\circ \)
- \( 56^\circ \)
- \( 90^\circ \)
- \( 96^\circ \)
#### Detailed Analysis
In the given diagram, you should note the following relationships:
1. Since arc \( AC \cong \) arc \( BC \) and the measure of arc \( BC = 152^\circ \), the measure of arc \( AC \) is also \( 152^\circ \).
2. The central angle subtended by arc \( BC \) which is angle \( \angle BOC \) would be \( 152^\circ \).
3. The angle formed outside the circle by a tangent and a secant (or two secants) from an external point is half the difference of the measures of the intercepted arcs.
Therefore, the measure of \( \angle D \) is:
\[ m\angle D = \frac{1}{2} \times ( \text{measure of larger arc} - \text{measure of smaller arc} ) \]
Since arc \( AC \cong \) arc \( BC \), the larger arc is the remaining arc, which is the entire circle minus both \( AC \) and \( BC \):
\[ \text{Measure of larger arc} = 360^\circ - 152^\circ - 152^\circ = 56^\circ \]
Now plugging in the values:
\[ m\angle D = \frac{1}{2} \times (56^\circ - 0^\circ) = \frac{1}{2} \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3539d619-7940-458b-b7f0-035dcdd97521%2F16a55f6f-c35a-4ad2-84d1-36eaf9004564%2F20lo5ic_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Problem on Tangents and Secants
#### Problem Statement
In the diagram below, tangent \( DA \) and secant \( DBC \) are drawn to circle \( O \) from external point \( D \), such that arc \( AC \cong \) arc \( BC \). If arc \( BC = 152^\circ \), determine and state \( m\angle D \).
#### Diagram Explanation
- Circle \( O \) has a center at point \( O \).
- Tangent \( DA \) touches the circle at point \( A \).
- Secant \( DBC \) intersects the circle at points \( B \) and \( C \).
- Arc \( AC \) is congruent to arc \( BC \).
- The measure of arc \( BC \) is \( 152^\circ \).
#### Multiple Choice Options
- \( 48^\circ \)
- \( 56^\circ \)
- \( 90^\circ \)
- \( 96^\circ \)
#### Detailed Analysis
In the given diagram, you should note the following relationships:
1. Since arc \( AC \cong \) arc \( BC \) and the measure of arc \( BC = 152^\circ \), the measure of arc \( AC \) is also \( 152^\circ \).
2. The central angle subtended by arc \( BC \) which is angle \( \angle BOC \) would be \( 152^\circ \).
3. The angle formed outside the circle by a tangent and a secant (or two secants) from an external point is half the difference of the measures of the intercepted arcs.
Therefore, the measure of \( \angle D \) is:
\[ m\angle D = \frac{1}{2} \times ( \text{measure of larger arc} - \text{measure of smaller arc} ) \]
Since arc \( AC \cong \) arc \( BC \), the larger arc is the remaining arc, which is the entire circle minus both \( AC \) and \( BC \):
\[ \text{Measure of larger arc} = 360^\circ - 152^\circ - 152^\circ = 56^\circ \]
Now plugging in the values:
\[ m\angle D = \frac{1}{2} \times (56^\circ - 0^\circ) = \frac{1}{2} \times
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