Rysunek pomocniczy:   

Zauważmy, że:

$\left|A{}^{\prime }B{}^{\prime }\right|=\left|AC\right|=10$  

Zatem:

$\frac{\left|AC\right|}{\left|AB\right|}=\cos \sphericalangle CAB$  

$\frac{10}{24}=\cos \sphericalangle CAB$  

$\frac{5}{12}=\cos \sphericalangle CAB$  

$0,4167\approx \cos \sphericalangle CAB$  

$\cos {65}^{\circ }\approx \cos \sphericalangle CAB$  

${65}^{\circ }\approx \left|\sphericalangle CAB\right|$  

 

Odp. B