Question
Six closed points are marked on a circle. Some of them are joined by line segments. The closed points are labeled upper A, upper B, upper C, upper D, upper E, and upper O.
The m∠DEC=27°
. Find the mCD
.
(1 point)
mCD=
°
The m∠DEC=27°
. Find the mCD
.
(1 point)
mCD=
°
Answers
GPT-4o mini
To find the measure of angle \( m\angle CD \) given that \( m\angle DEC = 27^\circ \), we can use the property of angles formed by chords intersecting inside a circle.
The property states that if you have two chords \( DE \) and \( CD \), the angle formed by these chords at the intersection is equal to half the sum of the measures of the arcs subtended by the angles. More specifically:
\[
m\angle CD = \frac{1}{2} (m\overset{\frown}{DE} + m\overset{\frown}{CE})
\]
However, we also know that:
\[
m\angle DEC = \frac{1}{2} (m\overset{\frown}{DC} + m\overset{\frown}{EC})
\]
Given that \( m\angle DEC = 27^\circ \), we can write:
\[
27^\circ = \frac{1}{2} (m\overset{\frown}{DC} + m\overset{\frown}{EC})
\]
Thus,
\[
m\overset{\frown}{DC} + m\overset{\frown}{EC} = 54^\circ
\]
Now, since the arcs in a circle sum to \( 360^\circ \), supposing \( m\overset{\frown}{DE} \) and arcs \( m\overset{\frown}{CD} + m\overset{\frown}{CE} + m\overset{\frown}{DE} = 360^\circ \).
To find \( m\angle CD \), we can rearrange the relationship:
If you consider \( m\angle DEO \) which is vertically opposite to \( m\angle CD \) and also sums up with \( m\overset{\frown}{DC} \) and \( m\overset{\frown}{EC} \), we could eventually find an equivalent angle to \( m\angle CD = 27^\circ \).
So thus we conclude:
\[
m\angle CD = 27^\circ
\]
Hence, \( mCD = \text{(The measure of angle CD) } = 27^\circ \).
The property states that if you have two chords \( DE \) and \( CD \), the angle formed by these chords at the intersection is equal to half the sum of the measures of the arcs subtended by the angles. More specifically:
\[
m\angle CD = \frac{1}{2} (m\overset{\frown}{DE} + m\overset{\frown}{CE})
\]
However, we also know that:
\[
m\angle DEC = \frac{1}{2} (m\overset{\frown}{DC} + m\overset{\frown}{EC})
\]
Given that \( m\angle DEC = 27^\circ \), we can write:
\[
27^\circ = \frac{1}{2} (m\overset{\frown}{DC} + m\overset{\frown}{EC})
\]
Thus,
\[
m\overset{\frown}{DC} + m\overset{\frown}{EC} = 54^\circ
\]
Now, since the arcs in a circle sum to \( 360^\circ \), supposing \( m\overset{\frown}{DE} \) and arcs \( m\overset{\frown}{CD} + m\overset{\frown}{CE} + m\overset{\frown}{DE} = 360^\circ \).
To find \( m\angle CD \), we can rearrange the relationship:
If you consider \( m\angle DEO \) which is vertically opposite to \( m\angle CD \) and also sums up with \( m\overset{\frown}{DC} \) and \( m\overset{\frown}{EC} \), we could eventually find an equivalent angle to \( m\angle CD = 27^\circ \).
So thus we conclude:
\[
m\angle CD = 27^\circ
\]
Hence, \( mCD = \text{(The measure of angle CD) } = 27^\circ \).