Given an equilateral triangle $ABC$ and a point $O$ on the plain of $\triangle ABC$, such that $\angle AOC=90^{\circ}, \angle BOC=75^{\circ}$. Find the angles of the triangle constructed from segments $AO,BO,CO$.
I have proof with trigonometry, but I am interested in a solution without it! Thanks in advance! (The answer is $135,30,15$.
So we have equilateral triangle $\triangle ABC$ with point $O$ such that $\angle AOC=90^{\circ}, \angle BOC=75^{\circ}$.
Construct equilateral triangle $\triangle BOP$:
$$\angle CBA=\angle PBO\implies \angle CBP=\angle ABO,$$ $$BP=BO,\space BC=BA$$
The conclusion is:
$$\triangle BPC \cong \triangle BOA$$
$$\angle CPB=\angle AOB=90^{\circ}+75^{\circ}=165^{\circ}$$
$$OA=PC$$
So the triangle $\triangle OPC$ has sides $OP=OB$, $PC=OA$ and $OC$. If we find the angles of that triangle, we are done. And that is easy:
$$\angle COP = \angle COB-\angle POB=75^{\circ}-60^{\circ}=15^{\circ}$$
$$\angle CPO=360^{\circ}-\angle CPB-\angle BPO=360^{\circ}-165^{\circ}-60^{\circ}=135^{\circ}$$
The third angle is:
$$\angle OCP=180^{\circ}-\angle COP-\angle CPO=180^{\circ}-135^{\circ}-15^{\circ}=30^{\circ}$$