Zauważmy, że trójkąty ADC i DBC są równoramienne


$\beta =\left|ACD\right|+\left|DCB\right|$  

$\alpha =\left|ACD\right|$  

$\left|BDC\right|=\left|DCB\right|$  

$\alpha +\alpha +\left|ADC\right|={180}^{\circ }$  

$\left|ADC\right|+\left|BDC\right|={180}^{\circ }$  


$a\text{)}\alpha ={10}^{\circ }$  

${10}^{\circ }+{10}^{\circ }+\left|ADC\right|={180}^{\circ }$  

${20}^{\circ }+\left|ADC\right|={180}^{\circ }\mid -{20}^{\circ }$  

$\left|ADC\right|={160}^{\circ }$  

zatem:

${160}^{\circ }+\left|BDC\right|={180}^{\circ }\mid -{160}^{\circ }$

$\left|BDC\right|={20}^{\circ }$   

$\beta ={10}^{\circ }+{20}^{\circ }={30}^{\circ }$  


$\alpha ={11}^{\circ }$  

${11}^{\circ }+{11}^{\circ }+\left|ADC\right|={180}^{\circ }$

${22}^{\circ }+\left|ADC\right|={180}^{\circ }\mid -{22}^{\circ }$  

$\left|ADC\right|={158}^{\circ }$  

zatem:

${158}^{\circ }+\left|BDC\right|={180}^{\circ }\mid -{158}^{\circ }$

$\left|BDC\right|={22}^{\circ }$   

$\beta ={11}^{\circ }+{22}^{\circ }={33}^{\circ }$  


$\alpha ={12}^{\circ }$  

${12}^{\circ }+{12}^{\circ }+\left|ADC\right|={180}^{\circ }$  

${24}^{\circ }+\left|ADC\right|={180}^{\circ }\mid -{24}^{\circ }$  

$\left|ADC\right|={156}^{\circ }$  

zatem:

${156}^{\circ }+\left|BDC\right|={180}^{\circ }\mid -{156}^{\circ }$  

$\left|BDC\right|={24}^{\circ }$   

$\beta ={12}^{\circ }+{24}^{\circ }={36}^{\circ }$  


$b\text{)}$  korzystając z  rozwiązanie podpunktu a) możemy domniemać, że

$\beta =3\alpha$  

sprawdźmy czy dla 

$\alpha ={13}^{\circ }$  

$\beta ={39}^{\circ }$  

sprawdzamy przypuszczenia za pomocą rachunków:

$\alpha ={13}^{\circ }$  

${13}^{\circ }+{13}^{\circ }+\left|ADC\right|={180}^{\circ }$  

${26}^{\circ }+\left|ADC\right|={180}^{\circ }\mid -{26}^{\circ }$  

$\left|ADC\right|={154}^{\circ }$  

zatem:

${154}^{\circ }+BDC={180}^{\circ }\mid -{154}^{\circ }$

  $\left|BDC\right|={26}^{\circ }$   

$\beta ={13}^{\circ }+{26}^{\circ }={39}^{\circ }$