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In the diagram shown, chords AB and CD intersect at E. The measure of (AC) ̂ is 120°, the measure of (DB) ̂ is (2x)° and the measure of ∠AEC is (4x)°. What is the degree measure of ∠ AED?

In the diagram shown chords AB and CD intersect at E The measure of AC is 120 the measure of DB is 2x and the measure of AEC is 4x What is the degree measure of class=

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Answer:

The measure of angle < AED is [tex]100\°[/tex]

Step-by-step explanation:

we know that

The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite.

so

[tex]m<AEC=\frac{1}{2}(arc\ AC+arc\ DB)[/tex]

substitute the values and solve for x

[tex]4x=\frac{1}{2}(120\°+2x)[/tex]

[tex]8x=(120\°+2x)[/tex]

[tex]8x-2x=120\°[/tex]

[tex]6x=120\°[/tex]

[tex]x=20\°[/tex]

[tex]m<AEC=4x=4(20\°)=80\°[/tex]

[tex]m<AEC+m<AED=180\°[/tex] -----> by supplementary angles

substitute value

[tex]80\°+m<AED=180\°[/tex]

[tex]m<AED=180\°-80\°=100\°[/tex]

You can use the fact that mean of opposite arc made by intersecting chord is measure of angle made by those intersecting line with each other which faces those arcs.

The degree measure of  ∠ AED is 100 degrees.

How to find measure of  ∠ AED?

For given figure. we have:

[tex]m\angle AEC = m\angle DEB = \dfrac{1}{2}(arc AC + arcDB) = 120 + 2x\\ 4x = \dfrac{1}{2}(120 + 2x)\\ 4x = 60 + x\\ 3x = 60\\ x = 20[/tex]

Thus, we have:

[tex]m\angle AEC = 4x = 4 \times 20 = 80^\circ[/tex]

Since angle AEC and AED add up to 180 degrees(since they make straight line), thus:

[tex]m\angle AEC + m\angle AED = 180^\circ\\ m\angle AED = 180 - m\angle AEC = 180 - 80 = 100^\circ[/tex]

Thus, we have measure of angle AED as:

[tex]m\angle AED = 100^\circ[/tex]

Learn more about arcs and angles in a circle here:

https://brainly.com/question/1364009

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