Answer:
wheres the picture?
Step-by-step explanation:
Answer:
The length of segment XY can be found by solving for a in

The measure of the central angle
is
.
Step-by-step explanation:
If the regular octagon has a perimeter of 122.4cm, then each side is 
The measure of each central angle is 
The angle between the apothem and the radius is 
The segment XY=a is the height of the right isosceles triangle.
We can use the Pythagoras Theorem with right triangle XYZ to get:


Therefore, the correct options are:
The length of segment XY can be found by solving for a in

The measure of the central angle
is
.
The measure of ∠ACB will be 110°
<u><em>Explanation</em></u>
According to the diagram below,
and
are the perpendicular bisectors of
and
respectively and they intersect side
at points
and
respectively.
So,
and 
Now, <u>according to the
postulate</u>, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other.
That means, ∠PCE = ∠PAE and ∠FCQ = ∠FBQ
As ∠CPQ = 78° , so ∠PCE + ∠PAE = 78° or, ∠PCE =
° and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ =
°
Now, in triangle CPQ, ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°
Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°
0 because 10 - 10 equals 0
We know that:

There is also an interesting property that relates the sine and the cosine of an angle:

We can find the cosine of theta using this equation:
![\begin{gathered} \cos ^2(\theta_1)=1-\sin ^2(\theta_1) \\ \cos (\theta_1)=\sqrt{1-\sin^2(\theta_1)} \\ \cos (\theta_1)=\sqrt[]{1-(-\frac{12}{13})^2} \\ \lvert\cos (\theta_1)\rvert=\sqrt[]{1-\frac{144}{169}}=\sqrt[]{\frac{25}{169}} \\ \lvert\cos (\theta_1)\rvert=\frac{5}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%20%5E2%28%5Ctheta_1%29%3D1-%5Csin%20%5E2%28%5Ctheta_1%29%20%5C%5C%20%5Ccos%20%28%5Ctheta_1%29%3D%5Csqrt%7B1-%5Csin%5E2%28%5Ctheta_1%29%7D%20%5C%5C%20%5Ccos%20%28%5Ctheta_1%29%3D%5Csqrt%5B%5D%7B1-%28-%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%20%5C%5C%20%5Clvert%5Ccos%20%28%5Ctheta_1%29%5Crvert%3D%5Csqrt%5B%5D%7B1-%5Cfrac%7B144%7D%7B169%7D%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B25%7D%7B169%7D%7D%20%5C%5C%20%5Clvert%5Ccos%20%28%5Ctheta_1%29%5Crvert%3D%5Cfrac%7B5%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Since theta is in the third quadrant then its cosine must be a negative number so: