Finding an angle in a figure with a regular octagon and regular pentagon

Question

I've recreated the diagram using Adobe Illustrator.

The pink polygon is a regular octagon and the grey is a regular pentagon.

I've tried to solve this for a while now but get nowhere. It seems like all I do is create more random angles that don't contribute to the solution. I've tried extending lines but that doesn't seem to help me at all.

Most other problems I've been given are relatively easy.

Answer 1

Let $y$ be the given angle (supposedly $67$, though it can't really be).

The vertex angle of a regular pentagon is $180-360/5 = 108$.
The vertex angle of a regular octagon is $180-360/8 = 135$.

The irregular pink pentagon on the right then has five angles which are clockwise from the top:
$x$, $135$, $135$, $135-108$, $360-108-y$.

The sum of the angles of a pentagon is $(5-2)*180 = 540$. Therefore we get that $x+135+135+135-108+360-108-y = 540$, or $x=y-9$.

If $y$ really were 67, then $x$ would indeed be $58$, but those numbers do not match the reality of the drawing. Those angles could be right if the regular pentagon had a smaller edge length, but still shared its bottom-right vertex with the octagon.

Answer 2

Hint:

The sum of the internal angles of a convex polygon with $n$ sides (the angles between two consecutive sides) is equal to $\; (n-2)\pi$ (in radians).

Answer 3

Assuming the common sides to be unit and setting the origin at the bottom left vertex, the angle $x$ is determined by its apex at $(0,1+\sqrt2)$ and the points $(1/\sqrt2,1+1/\sqrt2)$ and $(\cos2\pi/5,\sin2\pi/5)$.

The tangent of $x$ is given by the ratio of the cross-product of the two legs over their dot product.

$$x=\arctan\frac{\frac1{\sqrt2}\left(\frac{\sqrt{10+2\sqrt5}}4-1-\sqrt2\right)+\frac1{\sqrt2}\frac{\sqrt5-1}4}{\frac1{\sqrt2}\frac{\sqrt5-1}4-\frac1{\sqrt2}\left(\frac{\sqrt{10+2\sqrt5}}4-1-\sqrt2\right)}.$$

Answer 4

This problem can not be solved: the inner corners of a regular pentagon are $108°$. Draw a straigth line connecting the right points of the upper and lower edges of the regular octagon. This gives a triangle. The lower corner is $108° - 90° = 18°$. The corner formed by the inner corner of the regular pentagon and the $67°$ forms a total of $108° + 67° = 175°$ and the corners of a triangle only have a total sum of $180°$... (i.e. the problem is not well written).