In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline are diameters of circle PPP. What is the arc measure of \stackrel{\LARGE{\frown}}{CD} CD ⌢ C, D, start superscript, \frown, end superscript in degrees? ^\circ ∘ degrees

We know that center angle is equal to the arc length. Hence, the value of arc CD is equal to the angle CPD , which is 64 degree.

Given-

In the given fig (attached in the solution), we have,

AD and BE are the diameter of the circle P.

Center of the circle is P.

PC is the radius of the circle P.

APE is right angle triangle which is 90 degrees.

AD is a straight line and PE cut it into the center. Thus the value of angle APE is equal to the angle DPE. Hence the value of DPE is 90 degrees.

But in the given figure the value of DPE is,

equate both the values of the angle DPE, we get,

Now the value of angle CPD from the given fig is,

Put this value of k in the angle of CPD, we get,

As we know that center angle is equal to the arc length. Hence, the value of arc CD is equal to the angle CPD , which is 64 degree.

For more bout the arc, follow the link below-

link

The measure of the arc CD = 64°

Step-by-step explanation:

The rest of the question is the attched figure.

It is required to find the measure of the arc CD in degrees.

as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°

answer: i ! the answer is   9000 seconds, also d

hope i

-lavira : )