Rysunek poglądowy

Zauważmy, że zgodnie z twierdzeniem o dwusiecznych

$\frac{\left|CB\right|}{\left|AB\right|}=\frac{\left|CD\right|}{\left|DA\right|}$

$\frac{4}{8}=\frac{\left|CD\right|}{\left|DA\right|}$  

Zatem oznaczmy

$\left|CD\right|=x,\left|DA\right|=2x$  

Z twierdzenia cosinusów

${\left|AC\right|}^2={\left|AB\right|}^2+{\left|BC\right|}^2-2\cdot \left|AB\right|\cdot \left|BC\right|\cdot \cos \sphericalangle ABC$  

${\left|AC\right|}^2=8^2+4^2-2\cdot 8\cdot 4\cdot \cos {56}^{\circ }$  

${\left|AC\right|}^2=64+16-64\cdot \cos {56}^{\circ }$  

${\left|AC\right|}^2\approx 80-64\cdot 0,5592$

${\left|AC\right|}^2\approx 80-35,7888$

${\left|AC\right|}^2\approx 44.2112$  

$\left|AC\right|\approx 6,65$  

Zatem

$\left|AC\right|=3x=6,65$  

$\left|DA\right|=2x\approx 4,434$

Zachodzi stosunek pól

  $\frac{P_{ADB}}{P_{DCB}}=\frac{2}{1}$  

Zauważmy, że w takim razie

$P_{ADB}=\frac{2}{3}{P}_{ABC}$  

Skorzystamy w przypadku obu pól ze wzoru

 

$P=\frac{1}{2}bc\sin \alpha$

 

$P_{ADB}=\frac{1}{2}\cdot 8\cdot \left|BD\right|\sin {28}^{\circ }$  

$P_{ABC}=\frac{1}{2}\cdot 8\cdot 4\sin {56}^{\circ }$  

$\frac{1}{\cancel{2}}\cdot {\cancel{8}}^4\cdot \left|BD\right|\cdot \sin {28}^{\circ }=\frac{\cancel{2}}{3}\cdot \frac{1}{\cancel{2}}\cdot 8\cdot 4\cdot \sin {56}^{\circ }$  

$4\left|BD\right|\cdot \sin {28}^{\circ }=\frac{32}{3}\cdot \sin {56}^{\circ }\mid :4$  

$\left|BD\right|\cdot \sin {28}^{\circ }=\frac{8}{3}\cdot \sin {56}^{\circ }\mid :\sin {28}^{\circ }$  

$\left|BD\right|=\frac{8}{3}\cdot \sin {56}^{\circ }\cdot \frac{1}{\sin {28}^{\circ }}$  

$\left|BD\right|\approx \frac{8}{3}\cdot 0,829\cdot \frac{1}{0,4695}$   

$\left|BD\right|\approx \frac{6,632}{1,4085}$  

$\left|BD\right|\approx 4,71$