Answer:
[tex]arc\ DBE=206^o[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
The question is
Find the measure of arc DBE
step 1
Find the value of k
we know that
[tex]m\angle EPC=90^o[/tex] ----> by quarter of circle
[tex]m\angle EPC=(33k-9)^o[/tex]
so
[tex](33k-9)^o=90^o[/tex]
solve for k
[tex]33k=90+9\\33k=99\\k=3[/tex]
step 2
Find the measure of angle DPC
we know that
[tex]m\angle DPC=(20k+4)^o[/tex]
substitute the value of k
[tex]m\angle DPC=(20(3)+4)=64^o[/tex]
step 3
Find the measure of angle BPD
we know that
[tex]m\angle BPD+m\angle DPC=90^o[/tex] ---> by complementary angles
substitute the given value
[tex]m\angle BPD+64^o=90^o[/tex]
[tex]m\angle BPD=90^o-64^o=26^o[/tex]
step 4
Find the measure of arc DBE
we know that
[tex]arc\ DBE=arc\ DB+arc\ BAE[/tex]
[tex]arc\ DB=m\angle BPD=26^o[/tex] ----> by central angle
[tex]arc\ BAE=180^o[/tex] ---> because a diameter divide the circle into two equal parts (BE is a diameter)
substitute
[tex]arc\ DBE=26^o+180^o=206^o[/tex]
Answer:
In the figure below, \overline{AC} AC
start overline, A, C, end overline and \overline{BD} BD
start overline, B, D, end overline are diameters of circle PPP.
What is the arc measure of major arc \stackrel{\large{\frown}}{BDC}
BDC⌢ B, D, C, start superscript, \frown, end superscript in degrees?
the answer is 205
Step-by-step explanation:
