Rozwiązanie

 

 

 

 

 

$\left|\sphericalangle BAD\right|=180^{\circ }-90^{\circ }-35^{\circ }=180^{\circ }-125^{\circ }=55^{\circ }$

$\left|\sphericalangle ADC\right|=180^{\circ }-55^{\circ }=125^{\circ }$

$\left|\sphericalangle BDC\right|=125^{\circ }-90^{\circ }=35^{\circ }$

$\left|BC\right|=\left|CD\right|$, więc $\left|\sphericalangle BDC\right|=\left|\sphericalangle CBD\right|=35^{\circ }$

$\left|\sphericalangle BCD\right|=180^{\circ }-2\cdot 35^{\circ }=180^{\circ }-70^{\circ }=110^{\circ }$

Miara kąta $BCD$ jest równa $\stackrel{110^{\circ }}{{}_{...................}}$.