Rysunek pomocniczy:

Z treści zadania wiemy, że:

$\left|AC\right|=20\left[\text{cm}\right]\Rightarrow \left|AE\right|=\left|EC\right|=10\left[\text{cm}\right]$  

$\left|BD\right|=30\left[\text{cm}\right]$  

$\left|DE\right|=2x,\left|EB\right|=x$  

Zatem:

$2x+x=30\left[\text{cm}\right]$  

$3x=30\mid :3$  

$x=10\left[\text{cm}\right]$  

 

Zatem otrzymujemy:

$\mathrm{tg}\alpha =\frac{\left|AE\right|}{\left|ED\right|}$  

$\mathrm{tg}\alpha =\frac{10}{20}$  

$\mathrm{tg}\alpha =\frac{1}{2}$  

$\alpha \approx {27}^{\circ }$  

 

$\mathrm{tg}\beta =\frac{\left|DE\right|}{\left|AE\right|}$  

$\mathrm{tg}\beta =\frac{20}{10}$  

$\mathrm{tg}\beta =2$  

$\beta \approx {63}^{\circ }$  

 

$\mathrm{tg}\gamma =\frac{\left|EB\right|}{\left|AE\right|}$  

$\mathrm{tg}\gamma =\frac{10}{10}$  

$\mathrm{tg}\gamma =1$  

$\gamma ={45}^{\circ }$  

 

$\mathrm{tg}\delta =\frac{\left|AE\right|}{\left|EB\right|}$  

$\mathrm{tg}\delta =\frac{10}{10}$  

$\mathrm{tg}\delta =1$  

$\delta ={45}^{\circ }$  

 

Wyznaczmy kąty tego deltoidu.

$2\alpha \approx 2\cdot {27}^{\circ }={54}^{\circ }$  

$\beta +\gamma \approx {63}^{\circ }+{45}^{\circ }={108}^{\circ }$  

$2\delta -2\cdot {45}^{\circ }={90}^{\circ }$  

 

Odp. Kąty tego deltoidu to:  ${54}^{\circ },{108}^{\circ },{90}^{\circ },{108}^{\circ }.$