How to calculate 2 unknown angles of a equilateral non-equiangular pentagon given 3 known angles?

Question

I'm trying to figure out how to calculate 2 unknown angles of a equilateral non-equiangular pentagon given 3 known angles.

My intuition tells me there should only be one solution for the resulting 2 angles but I do not know how to work it out.

I've added this illustration of the problem.

Here is another variation of the problem with 2 known angles.

And here is another variation of the problem given 2 angles.

The 2 angle problems will have two solutions if the pentagon can go concave, but if we restrict it to a convex equalateral pentagon there should just be one solution. Right?

Answer 1

A direct brute force solution is to plug the whole thing in a coordinate system. Using your illustration:

As shown, you may calculate the next point by adding $(cos a, sin a)$ to each point, where $a$ is the angle between the side and positive $x$ axis.

As you arrive at the last point, say it is $(X,Y)$, then we simply have that angle $b$ is given by $\pi-asin Y$ or $acos X$, which should be the same. The other angle would thus be $3\pi$ minus all known angles.

However I need to point out that in this scheme the conditions, or constraints, could be too much instead of just enough; not all possible $3$ given angles would form a such a pentagon.

As shown in the picture below, if the angle in the lower right were not given, it would result in two solutions, since there are only two points with an equal given distance to both fixed end points (intersection of circles).

However, if a "wrong" angle (not giving one of the two solutions) was given, it cannot form a pentagon.

The other case is done in a similar manner. Notice that the centers of the two circles are fixed and known, and their radii are also fixed and known.

Answer 2

One way to solve it is to use the Law of cosines.

Let $\widehat{BCD}$ be the angle for the right point of your schema.

Then the law of cosines gives that the length of segment $[BD]$ verifies: $$BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot cos(\widehat{BCD})$$

Now that you have $BD$, you can use again the law of cosines to determine the angle $\widehat{CBD}$ which satifies: $$cos(\widehat{CBD}) = \frac{CB^2 + BD^2 - CD^2}{2 \cdot CB \cdot BD}$$

And from that angle you can derive $\widehat{ABD} = \widehat{ABC} - \widehat{CBD}$.

Then again, the law of cosines allows you to get a formula for the distance $AD$ from $AB$, $BD$ and $\widehat{ABD}$. Then you can get $\widehat{BAD}$ and $\widehat{EAD}$ (still using the law of cosines) because you know the length of the sides of triangles $ABD$ and $AED$.

Finally, you get $\widehat{BAE} = \widehat{BAD} + \widehat{EAD}$, which is one angle you are looking for.

In the same way, you can get the angle $\widehat{AED}$.

Answer 3

In the convex case, given that a pentagon with such angles exist, you can calculate the others with laws of cosine, sine and trig identities.

For example in the first picture I labeled the the top center point $A$ it's known angle $\alpha$ and so on in a clockwise manner. And called the sidelength $a$. Then the unknown $\delta$ angle can be calculated by first calculating $$\overline{AC} = \sqrt(2a^2(1-cos(\beta)))$$ Then calculating $$\overline{EC}^2 = 3a^2-2a^2cos(\beta)-2a^2sin(\alpha+\beta/2)\sqrt(2-2cos(\beta))$$ from triangle $ACE$ (with the cosine of angle $EAC$ being equal to $sin(\alpha+\beta/2))$, other than that it's really just algebra-autopilot) but from triangle $CDE$ This also equals $2a^2(1-cos(\delta))$. Subtracting $2a^2$ from bith sides and dividing by $2a^2$ you get: $$ cos(\delta) = \frac{-1+2cos(\beta)+2sin(\alpha+\beta/2)\sqrt(2-2cos(\beta))}{2}$$ Which is not that ugly as it could be :D. Because of symmetry you can swap variable $\alpha$ with $\gamma$ and $\delta$ with $\epsilon$ and get the equation for the other unknown angle.