parmenides51 wrote: The three sides of the quadrilateral are equal, the angles between them are equal, respectively $90^o$ and $150^o$. Find the smallest angle of this quadrilateral in degrees. $\color{blue}\boxed{\textbf{Answer: 45}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $ABCD$ be the quadrilateral where $AB=BC=CD=a, \angle ABC=90$ and $\angle BCD=150$ $\triangle ABC$ is isosceles $$\Rightarrow \angle BAC= \angle BCA =45$$$$\Rightarrow AC=a\sqrt{2}, \angle ACD=105$$$$\angle CAD=\alpha \Rightarrow \angle CDA=75-\alpha$$In $\triangle ACD:$ $$\frac{S_{\alpha}}{S_{75-\alpha}}=\frac{CD}{AC}=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}$$$$\Rightarrow S_{\alpha}\sqrt{2}=S_{75-\alpha}=S_{75}C_{\alpha}-C_{75}S_{\alpha}...(I)$$$$S_{75}=S_{30}C_{45}+C_{30}S_{45}=\frac{\sqrt{3}+1}{2\sqrt{2}} \Rightarrow C_{75}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$In $(I):$ $$\Rightarrow S_{\alpha}\sqrt{2}=(\frac{\sqrt{3}+1}{2\sqrt{2}})C_{\alpha}-(\frac{\sqrt{3}-1}{2\sqrt{2}})S_{\alpha}$$$$\Rightarrow 2S_{\alpha}=(\frac{\sqrt{3}+1}{2})C_{\alpha}-(\frac{\sqrt{3}-1}{2})S_{\alpha}$$$$\Rightarrow (\frac{\sqrt{3}+3}{2})S_{\alpha}=(\frac{\sqrt{3}+1}{2})C_{\alpha}$$$$\Rightarrow \frac{C_{\alpha}}{S_{\alpha}}=\sqrt{3}$$$$\Rightarrow cot\alpha=\sqrt{3}, \alpha \in [ 0,75]$$$$\Rightarrow \alpha=30$$$$\Rightarrow \text{The angles of the quadrilateral are } 90, 150, 75 \text{ and } 45$$$$\Rightarrow \boxed{\textbf{The smallest angle is } 45}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

parmenides51 wrote: The three sides of the quadrilateral are equal, the angles between them are equal, respectively $90^o$ and $150^o$. Find the smallest angle of this quadrilateral in degrees.

See the drawing & explanation. Best regards, sunken rock