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The figure below is Circle E. Line CF is tangent at point C.

Find the measure of Angle ECF.
Find the measure of Angle AKB.
Find the measure of Angle ACF.

The figure below is Circle E Line CF is tangent at point C Find the measure of Angle ECF Find the measure of Angle AKB Find the measure of Angle ACF class=

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Answer:

Part 1) [tex]m<ECF=90\°[/tex]

Part 2) [tex]m<AKB=77.5\°[/tex]

Part 3) [tex]m<ACF=75\°[/tex]

Step-by-step explanation:

Part 1) Find the measure of angle ECF

we know that

CF is tangent at point C

so

the radius EC is perpendicular to the tangent CF

therefore

[tex]m<ECF=90\°[/tex]

Part 2) Find the measure of angle AKB

we know that

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

[tex]m<AKB=\frac{1}{2}(arc\ AB+arc\ DC)[/tex]

substitute the values

[tex]m<AKB=\frac{1}{2}(50\°+105\°)=77.5\°[/tex]

Part 3) Find the measure of angle ACF

we know that

The inscribed angle is half that of the arc it comprises

[tex]m<ACF=\frac{1}{2}(arc\ AB+arc\ BF)[/tex]

substitute the values

[tex]m<ACF=\frac{1}{2}(50\°+100\°)=75\°[/tex]

Answer:

∠ ECF = 90°

∠AKB = 77.5°

∠ACF = 75°

Step-by-step explanation:

The given figure is a circle E.

Line CF is tangent at point C.

(1) To find measures of angle ECF

Since line FC is tangent at point C and radius ECis always perpendicular to the tengent.

So ∠ ECF = 90°

(2) Measure of angle AKB

By theorem of angle formed by two interesting chords.

∠ AKB = [tex]\frac{1}{2}[/tex] (arc AB + arc CD)

= [tex]\frac{1}{2}[/tex] (50 + 105) = [tex]\frac{1}{2}(155)[/tex]

∠AKB = 77.5°

(3) Measure of angle ACF

Since tangent chord angle = [tex]\frac{1}{2}[/tex] (intercepted arc)

m (∠ACF) = [tex]\frac{1}{2}[/tex] (m arc AB + m arc BC)

= [tex]\frac{1}{2}(50+100)[/tex]

[tex]\frac{1}{2}(150)[/tex]

∠ACF = 75°

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