Answer:
Part a) The measure of angle BDC is [tex]m<BDC=48\°[/tex]
Part b) The measure of angle CPD is [tex]m<CPD=68\°[/tex]
Part c) The measure of arc AD is [tex]arc\ AD=128\°[/tex]
Part d) The measure of arc AB is [tex]arc\ AB=52\°[/tex]
Part e) The measure of arc CAD is [tex]arc\ CAD=276\°[/tex]
Part f) The measure of angle ABD is [tex]m<ABD=64\°[/tex]
Step-by-step explanation:
Part a) Find the measure of angle BDC
we know that
The inscribed angle is half that of the arc it comprises.
[tex]m<BDC=\frac{1}{2}(arc\ BC)[/tex]
substitute the value
[tex]m<BDC=\frac{1}{2}(96\°)=48\°[/tex]
Part b) Find the measure of angle CPD
we know that
The sum of the internal angles of a triangle must be equal to 180 degrees.
so In the triangle CPD
[tex]64\°+m<CPD+m<BDC=180\°[/tex]
substitute the values
[tex]64\°+m<CPD+48\°=180\°[/tex]
[tex]m<CPD=180\°-112\°=68\°[/tex]
Part c) Find the measure of arc AD
we know that
The inscribed angle is half that of the arc it comprises.
[tex]m<PCD=\frac{1}{2}(arc\ AD)[/tex]
substitute the values
[tex]64\°=\frac{1}{2}(arc\ AD)[/tex]
[tex]arc\ AD=128\°[/tex]
Part d) Find the measure of arc AB
Remember that BD is a diameter
so
[tex]arc\ AB+arc\ AD=180\°[/tex] ----> the diameter divide the circle into two equal parts
[tex]arc\ AB+128\°=180\°[/tex]
[tex]arc\ AB=180\°-128\°=52\°[/tex]
Part e) Find the measure of arc CAD
we know that
[tex]arc\ CAD=arc\ CB+arc\ BAD[/tex]
substitute the values
[tex]arc\ CAD=96\°+180\°=276\°[/tex]
Part f) Find the measure of angle ABD
we know that
The inscribed angle is half that of the arc it comprises.
[tex]m<ABD=\frac{1}{2}(arc\ AD)[/tex]
substitute the values
[tex]m<ABD=\frac{1}{2}(128\°)=64\°[/tex]
