Skorzystamy z  twierdzenia sinusów

$\frac{\left|AB\right|}{\sin \sphericalangle BCA}=\frac{\left|BC\right|}{\sin \sphericalangle CAB}$  

$\frac{24}{\sin {60}^{\circ }}=\frac{16}{\sin \sphericalangle CAB}\mid \cdot \sin \sphericalangle CAB$

$\frac{24}{\sin {60}^{\circ }}\cdot \sin \sphericalangle CAB=16\mid \cdot \sin {60}^{\circ }$

$24\cdot \sin \sphericalangle CAB=16\sin {60}^{\circ }\mid :24$

$\sin \sphericalangle CAB=\frac{16}{24}\sin {60}^{\circ }$

$\sin \sphericalangle CAB=\frac{\cancel{2}}{3}\cdot \frac{\sqrt{3}}{\cancel{2}}$

$\sin \sphericalangle CAB=\frac{\sqrt{3}}{3}$

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Prawidłowa odpowiedź to C.