Question

In the figure below, \overline{AC} AC start overline, A, C, end overline and \overline{BE} BE start overline, B, E, end overline are diameters of circle PPP. What is the arc measure of \stackrel{\Huge{\frown}}{DBE} DBE ⌢ D, B, E, start superscript, \frown, end superscript in degrees?

Answer 1

Answer:

$arc\ DBE=206^o$

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

$m\angle EPC=90^o$ ----> by quarter of circle

$m\angle EPC=(33k-9)^o$

so

$(33k-9)^o=90^o$

solve for k

$33k=90+9\\33k=99\\k=3$

step 2

Find the measure of angle DPC

we know that

$m\angle DPC=(20k+4)^o$

substitute the value of k

$m\angle DPC=(20(3)+4)=64^o$

step 3

Find the measure of angle BPD

we know that

$m\angle BPD+m\angle DPC=90^o$ ---> by complementary angles

substitute the given value

$m\angle BPD+64^o=90^o$

$m\angle BPD=90^o-64^o=26^o$

step 4

Find the measure of arc DBE

we know that

$arc\ DBE=arc\ DB+arc\ BAE$

$arc\ DB=m\angle BPD=26^o$ ----> by central angle

$arc\ BAE=180^o$ ---> because a diameter divide the circle into two equal parts (BE is a diameter)

substitute

$arc\ DBE=26^o+180^o=206^o$