a)

Zauważmy, że jest to trójkąt prostokątny równoramienny.

$\left|AB\right|=4$  

$\left|AC\right|=4\sqrt{2}$  

 

$P=\frac{1}{2}\cdot 4\cdot 4$  

$P=8$  

 

$r=\frac{2P}{a+b+c}=\frac{2\cdot 8}{4+4+4\sqrt{2}}=\frac{16}{8+4\sqrt{2}}=\frac{4}{2+\sqrt{2}}=\frac{4\left(2-\sqrt{2}\right)}{2^2-{\sqrt{2}}^2}=$ $r=\frac{2P}{a+b+c}=\frac{2\cdot 8}{4+4+4\sqrt{2}}=\frac{16}{8+4\sqrt{2}}=\frac{4}{2+\sqrt{2}}=\frac{4\left(2-\sqrt{2}\right)}{2^2-{\sqrt{2}}^2}=$

  $=\frac{8-4\sqrt{2}}{4-2}=\frac{8-4\sqrt{2}}{2}=4-2\sqrt{2}$  

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b)

$\frac{5}{\left|AC\right|}=\sin {30}^o$  

$5=\sin {30}^o\cdot \left|AC\right|$  

$5=\frac{1}{2}\cdot \left|AC\right|\mid \cdot 2$  

$10=\left|AC\right|$  

 

$\frac{5}{\left|AB\right|}=\mathrm{tg}{30}^o$  

$5=\mathrm{tg}{30}^o\cdot \left|AB\right|$  

$5=\frac{\sqrt{3}}{3}\cdot \left|AB\right|\mid :\frac{\sqrt{3}}{3}$  

$5\cdot \frac{3}{\sqrt{3}}=\left|AB\right|$  

$\frac{15\sqrt{3}}{3}=\left|AB\right|$  

$5\sqrt{3}=\left|AB\right|$  

 

$P=\frac{1}{2}\cdot 5\cdot 5\sqrt{3}=\frac{25}{2}\sqrt{3}$  

 

$r=\frac{2P}{a+b+c}=\frac{2\cdot \frac{25}{2}\sqrt{3}}{5+10+5\sqrt{3}}=\frac{25\sqrt{3}}{15+5\sqrt{3}}=\frac{5\sqrt{3}}{3+\sqrt{3}}=$  

$=\frac{5\sqrt{3}\left(3-\sqrt{3}\right)}{3^2-{\sqrt{3}}^2}=\frac{15\sqrt{3}-15}{9-3}=\frac{15\sqrt{3}-15}{6}=\frac{5\sqrt{3}-5}{2}$