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Skorzystamy z twierdzenia cosinusów

${\left|AC\right|}^2={11}^2+8^2-2\cdot 11\cdot 8\cdot \cos {60}^{\circ }$  

${\left|AC\right|}^2=121+64-2\cdot 88\cdot \frac{1}{2}$

${\left|AC\right|}^2=185-88$

${\left|AC\right|}^2=97$

$\left|AC\right|=\sqrt{97}$

Wiemy, że

$\left|\sphericalangle BCD\right|={180}^{\circ }-{60}^{\circ }={120}^{\circ }$  

Ponownie z twierdzenia cosinusów

${\left|BD\right|}^2={11}^2+8^2-2\cdot 11\cdot 8\cdot \cos {120}^{\circ }$  

${\left|BD\right|}^2=121+64-2\cdot 88\cdot \cos \left({180}^{\circ }-{60}^{\circ }\right)$  

${\left|BD\right|}^2=185-2\cdot 88\cdot \left(-\cos {60}^{\circ }\right)$

${\left|BD\right|}^2=185+2\cdot 88\cdot \frac{1}{2}$

${\left|BD\right|}^2=185+88$

${\left|BD\right|}^2=273$

$\left|BD\right|=\sqrt{273}$