$\cos \alpha =\frac{\frac{1}{2}\cdot \left|AB\right|}{30}$  

$\cos \alpha =\frac{\left|AB\right|}{60}$  

$\left|AB\right|=\cos \alpha \cdot 60=0,6\cdot 60=36$  

${\left|OA\right|}^2={\left|OD\right|}^2+{\left(\frac{1}{2}\left|AB\right|\right)}^2$  

$900={\left|OD\right|}^2+324$  

${\left|OD\right|}^2=576$  

$\left|OD\right|=24$  

$\left|CD\right|=r+\left|OD\right|=30+24=54$  

${\left|AC\right|}^2={\left|AD\right|}^2+{\left|CD\right|}^2$  

${\left|AC\right|}^2=324+2916$  

${\left|AC\right|}^2=3240$  

$\left|AC\right|=\sqrt{3240}=\sqrt{324\cdot 10}=18\sqrt{10}$  

Obwód:

$\text{Obw}=18\sqrt{10}+18\sqrt{10}+24=36\sqrt{10}+24$  

Pole:

$P=\frac{1}{2}\cdot \left|AB\right|\cdot \left|CD\right|=\frac{1}{2}\cdot 36\cdot 54=18\cdot 54=972$  

$tg\left(\angle CAD\right)=\frac{\left|CD\right|}{\left|AD\right|}=\frac{54}{18}=3$  

Znajdźmy najbliższą wartość dla tangensa kąta by oszacować kąt.

$tg{72}^o\approx 3,0777$  

$\angle CAD\approx {72}^o$

Kąty wewnętrzne to:

${72}^o,{72}^o,{36}^o$  

II przypadek:

$\cos \alpha =\frac{\frac{1}{2}\left|AB\right|}{30}$  

$\frac{\left|AB\right|}{60}=\cos \alpha$  

$\left|AB\right|=60\cdot \cos \alpha =60\cdot 0,6=36$  

Z twierdzenia Pitagorasa:

${\left(\frac{1}{2}\left|AD\right|\right)}^2+{\left|OD\right|}^2={\left|AO\right|}^2$  

$324+{\left|OD\right|}^2=900$  

$\left|OD\right|=24$  

$\left|CD\right|=\left|OC\right|-\left|OD\right|=30-24=6$   

Z twierdzenia Pitagorasa dla trójkąta ADC:

${\left|AC\right|}^2={\left|CD\right|}^2+{\left(\frac{1}{2}\left|AB\right|\right)}^2=36+324=360$  

$\left|AC\right|=\sqrt{360}=\sqrt{36\cdot 10}=6\sqrt{10}$  

Obwód:

$\text{Obw}=2\cdot \left|AC\right|+\left|AB\right|=12\sqrt{10}+36$  

Pole:

$P=\frac{1}{2}\cdot \left|AB\right|\cdot \left|CD\right|=\frac{1}{2}\cdot 36\cdot 6=108$  

Kąty wewnętrzne:

$tg\left(\angle CAD\right)=\frac{\left|CD\right|}{\left|AD\right|}=\frac{6}{18}=\frac{1}{3}\approx 0,3333$  

$\angle CAD\approx {19}^o$  

${19}^o,{19}^o,{142}^o$