The geometric construction of the $90^\circ, 87^\circ, 3^\circ$ triangle

Question

The construction of the $90^\circ, 45^\circ, 45^\circ$ and the $90^\circ, 60^\circ, 30^\circ$ triangles is well known.

How can be constructed a triangle with angles $90^\circ, 87^\circ, 3^\circ$ without using regular polygons? By "construction" I mean the determination of the proportion of its sides and the required internal angles by the common operations in plane geometry, like in the other triangles.

Is it possible to use the exterior angle theorem and the triangles $90^\circ, 72^\circ, 18^\circ$ and $90^\circ, 75^\circ, 15^\circ$ to construct geometrically the angle $18^\circ - 15^\circ= 3^\circ$? How this construction can be done?

Answer 1

A hint: $3^\circ=45^\circ+30^\circ-72^\circ$. For $72^\circ$ however you have to know how to construct a regular pentagon or some golden ratio triangle.

Answer 2

Following E.Girgin suggestion plus Richmond's method:

Steps:

1. Construct a regular hexagon $ABCDEF$ with centre $O$;
2. Construct a square in such a way that $\widehat{AOG}=75^\circ$;
   let $B'=BC\cap OG$ and take $A'\in FA$ such that $OA'\perp OB'$;
3. Let $\Gamma$ be the circle with centre $O$ through $A'$ and $M$ be the midpoint of $OA'$;
   let $N$ be the intersection between $OB'$ and the angle bisector of $\widehat{OMB'}$;
4. Let $A''$ be the point of intersection between $\Gamma$ and the parallel to $OA'$ through $N$;
5. $\color{red}{\widehat{AOA''}=3^\circ}$.

An equivalent construction through $3=\frac{72-60}{2\cdot 2}$:

Steps:

1. Let $AO=OB$ and $OA\perp OB$; let $\Gamma$ be the circle centered at $O$ through $A$;
2. Let $U,V,M$ be the midpoints of $BM,BO,OA$; let $W$ be the intersection between $UV$
   and the circumcircle of $BOM$, whose centre is clearly $U$;
3. Let $N=WM\cap OB$ and take $P\in\Gamma$ such that $NP\parallel OA$;
4. Take $Q\in\Gamma$ such that $QP=PO$;
5. By bisecting twice $\widehat{QOB}$, we get an angle with amplitude $\color{red}{3^\circ}.$