How do I calculate certain angles determined by sides and diagonals of a regular octagon? I have this exercise:

My development was:
$\angle{\beta} = 135°$, because this is a regular polygon.
$\angle{y} = 135° - 22.5° = 112.5°$, because the triangle $AGH$ is isosceles, then its equal angles are $22.5$, then $y + 22.5° = 135°$.
But I could not find the $\angle{\alpha} $. How could I do it? Thanks in advance.
 A: Also, if you inscribe the octagon into a circle, $\angle{\alpha}$ is the angle in a semi circle  $= 90^\circ$

A: In a regular $n$-gon the internal angle is  $(n-2)\times\frac{180^{\circ}}{n}$. For the regular octagon this is then $(8-2)\times\frac{180^{\circ}}{8}=6\times\frac{180^{\circ}}{8}=135^{\circ}$.
Hence $\beta = 135^{\circ}$. Now since $\triangle AHG$ is isosceles, then $\angle HAG=\angle HGA=22.5^{\circ}$. Therefore the ratio of the angles $\beta$, $\gamma$, $\alpha$ is 
$$\beta:\gamma:\alpha=135:135-22.5:135-2\times 22.5=135:112.5:90=6:5:4$$
So $\alpha+\beta+\gamma=90^{\circ}+135^{\circ}+112.5^{\circ}=337.5^{\circ}$.
In general if you dissect a regular polygon into triangles in the manner of the question the ratio of the angles is as
$$(n-2):(n-3):\dots:2:1$$
A: Any regular polygon of any number of sides can be inscribed in a circle.
The figure below shows what might be a few of the vertices 
($A$, $B$, $C$, $D$, and $E$) 
and sides of a regular polygon inscribed in a circle.

Now if these are vertices of a regular $n$-gon, we can continue the sequence around the circle until we get back to $A.$
Then we will have $n$ equal sides of the polygon, and together with the center of the circle each side forms an isoceles triangle
($\triangle AOB$, $\triangle BOC$, $\triangle COD$, etc.).
These triangles are all congruent, and their angles at $O$ sum to $360$ degrees, so each of these angles is $\frac{360}{n}$ degrees.
But the inscribed angle subtended by two points on a circle
is half as large as the central angle subtended by those same points;
that is, $\angle BAC$ is half as large as $\angle BOC,$
or to put it another way, $\angle BOC = 2\angle BAC.$
If we let $\theta = \angle BAC,$ then $\angle BOC = 2\theta,$
as shown in the figure.
But $\angle BOC$ is one of the vertex angles of the $n$ congruent isoceles triangles, that is,
$$ 2\theta = \angle BOC = \left(\frac{360}{n}\right)^\circ,$$
so $\theta = \left(\frac{180}{n}\right)^\circ.$
Since the central angles $\angle BOC$, $\angle COD$, $\angle DOE$, and so forth are all equal, so are the inscribed angles $\angle BAC$, $\angle CAD$, $\angle DAE$, and so forth.
That is,
$$ \angle BAC = \angle CAD = \angle DAE = \cdots = \theta. $$
It follows that
\begin{align}
 \angle BAD &= \angle BAC + \angle CAD = 2\theta, \\
 \angle BAE &= \angle BAC + \angle CAD + \angle DAE = 3\theta,
\end{align}
and so forth.
In other words, if the inscribed angle at $A$ subtends $k$ sides of the polygon, the angle is $k\theta.$
In a regular octagon, $n = 8$ (eight sides), so
$\theta = \left(\frac{180}{8}\right)^\circ = 22.5^\circ.$
In your figure, $\beta$ subtends six sides of the octagon,
$\gamma$ subtends five sides, and $\alpha$ subtends four sides, so
\begin{align}
\beta  &= (n-2)\theta = 6\theta = 135^\circ, \\
\gamma &= (n-3)\theta = 5\theta = 112.5^\circ, \\
\alpha &= (n-4)\theta = 4\theta = 90^\circ.
\end{align}
If you had drawn a regular polygon with more than eight sides,
and drawn segments from vertex $A$ to each other vertex,
the angles $\beta,$ $\gamma,$ and $\alpha$ formed in this way would
still measure $(n-2)\theta,$ $(n-3)\theta,$ $(n-4)\theta,$ and so forth.
A: You should notice that the inscribed angle is half of the centra angle opposite to the same arc.
Thus, $$\alpha\overset{m}=\frac{1}{2}\widehat {ABCDE}=\frac{1}{2}\cdot 360^o \cdot \frac{4}{8};$$
$$\beta\overset{m}=\frac{1}{2}\widehat {ABCDEFG}=\frac{1}{2}\cdot 360^o \cdot \frac{6}{8};$$
$$\gamma\overset{m}=\frac{1}{2}\widehat {ABCDEF}=\frac{1}{2}\cdot 360^o \cdot \frac{5}{8}.$$
As a result, $$\alpha+\beta+\gamma=\frac{1}{2}\cdot 360^o \cdot\frac{4+6+5}{8}=337.5^o.$$
A: 
Because $\angle AHG = \angle FGH$ and $AH = FG$, then $AHGF$ is an isosceles trapezoid with $AF$ parallel to $HG$. Note that$$
β = \angle AHG = \angle HGF = \angle GFE,
$$
thus$$
\angle AFG = π - \angle HGF = π - β\\
\Longrightarrow α = \angle AFE = \angle GFE - \angle AFG = 2β - π = \frac{π}{2}.
$$
A: Since the octagon is regular it can be inscribed in a circle as following

We can see that $\alpha$ is opposite of the diameter of the circle so is $90\text{ degrees}$ ans the summation would be$$\LARGE \alpha+\beta+\gamma=337.5\text{ degrees}$$
(picture from http://etc.usf.edu/clipart/43400/43451/8c_43451.htm with some modifications)
A: First: $\angle{GFA}=135^\circ-a.$ $\angle{GAF}=180^\circ-135^\circ+a-112.5^\circ=-67.5+a.$  (interior angle sum & triangle angle usm)
 $\angle{HAF}=\angle{HAG}+\angle{GAF}$ (angle addition property) is equal to $45^\circ.$ Attached is an image of an octagon that might help you believe the fact.
If we draw in a square, as shown, it is evident $\angle{HAF}$ is $45^\circ.$

These triangles are congruent by SSS (all sides are the same). Therefore, the corresponding angles are the same. This means twice $\angle{HAF}=90^\circ,$ implying $\angle{HAF}=45^\circ.$
$\angle{HAG}=\frac{180^\circ-135^\circ}2=22.5^\circ$ (isosceles property & triangle angle sum).
Therefore, $-67.5^\circ+a+22.5^\circ=45^\circ$ which leads to $a=90^\circ.$ 
A: Bear in mind just five basic things in filling up all angles.


*

*All segments subtend $\pi/8= 22.5^{\circ}$ at A (equal segments subtend equal agle at center. Corresponding angle subtended at boundary is half that.

*Every external angle is  $\pi/4= 45^{\circ}$ at each vertex $ = 2 \pi/8$rotation between sides

*Sum of three angles in triangle is $\pi= 180^{\circ}$

*Complementary angles sum to $\pi/4= 45^{\circ}$

*Angle in semi-circle $\pi/2= 90^{\circ}$
Required angle sum 
$$ 337.5^{\circ}  $$

