Answer:
Part 1) [tex]m\ arc\ CB=64\°[/tex]
Part 2) [tex]m<AOC=116\°[/tex]
Part 3) [tex]m\ arc\ AD=26\°[/tex]
Part 4) [tex]m\ arc\ DFB=154\°[/tex]
Part 5) [tex]m<CDB=32\°[/tex]
Step-by-step explanation:
Part 1) Find the measure of arc CB
we know that
[tex]m\ arc\ CB=m<COB[/tex] -----> by central angle
we have
[tex]m<COB=64\°[/tex] ----> given problem
so
[tex]m\ arc\ CB=64\°[/tex]
Part 2) Find the measure of angle AOC
we know that
[tex]m<AOC+m<COB=180\°[/tex] ------> by supplementary angles
we have
[tex]m<COB=64\°[/tex]
substitute
[tex]m<AOC+64\°=180\°[/tex]
[tex]m<AOC=180\°-64\°=116\°[/tex]
Part 3) Find the measure of arc AD
we know that
The inscribed angle measures half that of the arc comprising
so
[tex]m<ABD=\frac{1}{2}(m\ arc\ AD)[/tex]
we have
[tex]m<ABD=13\°[/tex]
substitute
[tex]13\°=\frac{1}{2}(m\ arc\ AD)[/tex]
[tex]m\ arc\ AD=26\°[/tex]
Part 4) Find the measure of arc DFB
we know that
[tex]m\ arc\ DFB=360\°-(m\ arc\ AD+m\ arc\ ACB)[/tex]
we have
[tex]m\ arc\ AD=26\°[/tex]
[tex]m\ arc\ ACB=180\°[/tex] -----> because is a diameter
substitute
[tex]m\ arc\ DFB=360\°-(26\°+180\°)=154\°[/tex]
Part 5) Find the measure of angle CDB
we know that
The inscribed angle measures half that of the arc comprising
so
[tex]m<CDB=\frac{1}{2}(m\ arc\ CB)[/tex]
we have
[tex]m\ arc\ CB=64\°[/tex]
substitute
[tex]m<CDB=\frac{1}{2}(64\°)=32\°[/tex]
